Sequence and Series Question Answers: NCERT Class 11 Mathematics

Welcome to the Chapter 9 - Sequence and Series, Class 11 Mathematics - NCERT Solutions page. Here, we provide detailed question answers for Chapter 9 - Sequence and Series.The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics and excel in their exams. By going through these Sequence and Series question answers, you can strengthen your foundation and improve your performance in Class 11 Mathematics. Whether you're revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Sequence is a collection of objects taken one by one in which repetitions may present but order matters. It can have any number of terms. When these terms are added it is called a series. We get some general expressions to solve sequences and series. It solves very tedious calculations which are very difficult to solve manually. This chapter consists of arithmetic progression and mean, geometric progression and mean, general terms, sum of n terms, arithmetic and geometric series, infinite G.P, relation between A.M and G.M, some special series.

Download PDF - Chapter 9 Sequence and Series - Class 11 Mathematics

Download PDF - NCERT Examplar Solutions - Chapter 9 Sequence and Series - Class 11 Mathematics

Exercise 1

Q1	Write the first five terms of the sequences whose nth term is a_n=n(n+2)
Ans:	a_n = n(n+2) Substituting n = 1, 2, 3, 4, and 5, we obtain a₁ = 1(1+2) = 3 a₂ = 2(2+2) = 8 a₃ = 3(3+2) = 15 a₄ = 4(4+2) = 24 a₅ = 5(5+2) = 35 Therefore, the required terms are 3, 8, 15, 24, and 35.

Q2	Write the first five terms of the sequences whose nth term is \begin{align} a_n = \frac {n}{n+1}\end{align}
Ans:	\begin{align} a_n = \frac {n}{n+1}\end{align} Substituting n = 1, 2, 3, 4, 5, we obtain \begin{align} a_1 = \frac {1}{1+1}=\frac{1}{2},a_2 = \frac {2}{2+1}=\frac{2}{3},a_3 = \frac {3}{3+1}=\frac{3}{4},a_4 = \frac {4}{4+1}=\frac{4}{5},a_5 = \frac {5}{5+1}=\frac{5}{6}\end{align} Therefore, the required terms are \begin{align} \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},and \frac{5}{6}\end{align}

Exercise 2

Q1	Find the sum of odd integers from 1 to 2001.
Ans:	The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001. This sequence forms an A.P. Here, first term, a = 1 Common difference, d = 2 Here, \begin{align} a + (n - 1)d = 2001 \end{align} \begin{align} => 1 + (n - 1)(2) = 2001 \end{align} \begin{align} => 2n -2 = 2000 \end{align} \begin{align} => n = 1001 \end{align} \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align} \begin{align} \therefore S_n = \frac {1001}{2}\left[2 × 1 + (1001 -1)×2\right]\end{align} \begin{align} = \frac {1001}{2}\left[2 + 1000×2\right]\end{align} \begin{align} = \frac {1001}{2} × 2002\end{align} \begin{align} =1001 × 1001 \end{align} \begin{align} = 1002001 \end{align} Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Q2	Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Ans:	The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995. This sequence forms an A.P. Here, first term, a = 105 Common difference, d = 5 Here, \begin{align} a + (n - 1)d = 995 \end{align} \begin{align} => 105 + (n - 1)5 = 995 \end{align} \begin{align} => (n - 1)5 = 995 - 105 = 890 \end{align} \begin{align} => n -1 = 178 \end{align} \begin{align} => n = 179 \end{align} \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align} \begin{align} \therefore S_n = \frac {179}{2}\left[2 × (105) + (179 -1)×(5)\right]\end{align} \begin{align} = \frac {179}{2}\left[2(105) + (178)(5)\right]\end{align} \begin{align} = 179\left[105 + (89)5\right]\end{align} \begin{align} = (179)\left[105 + 445\right]\end{align} \begin{align} =179 × 550 \end{align} \begin{align} = 98450 \end{align} Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Q3	In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Ans:	First term = 2 Let d be the common difference of the A.P. Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, … Sum of first five terms = 10 + 10d Sum of next five terms = 10 + 35d According to the given condition, \begin{align} => 10 + 10d = \frac{1}{4}(10 + 35d) \end{align} \begin{align} => 40 + 40d = 10 + 35d \end{align} \begin{align} => 30 = - 5d \end{align} \begin{align} => d = - 6 \end{align} \begin{align} \therefore a_{20} = a + (20 -1)d = 2 + (19)(-6) = 2 - 114 = -112\end{align} Thus, the 20th term of the A.P. is –112.

Q4	How many terms of the A.P. \begin{align} -6, -\frac{11}{2}, -5, ... \end{align} are needed to give the sum –25?
Ans:	Let the sum of n terms of the given A.P. be –25. It is known that, \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}, where n = number of terms, a = first term, and d = common difference Here, a = –6 \begin{align} d = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2}\end{align} Therefore, we obtain \begin{align} -25 = \frac {n}{2}\left[2 × (-6) + (n -1)×\frac{1}{2}\right]\end{align} \begin{align} => -50 = n\left[-12 + \frac{n}{2} - \frac{1}{2}\right]\end{align} \begin{align} => -50 = n\left[-\frac{25}{2} + \frac{n}{2}\right]\end{align} \begin{align} => -100 = n\left(-25 + n\right)\end{align} \begin{align} => n^2 - 25n + 100 = 0\end{align} \begin{align} => n^2 - 5n -20n + 100 = 0\end{align} \begin{align} => n(n - 5)- 20(n - 5) = 0\end{align} \begin{align} => n = 20 \; or\; 5\end{align}

Q5	In an A.P., if pth term is \begin{align} \frac{1}{q} \; and \;qth\; term \; is\; \frac{1}{p}\end{align} , prove that the sum of first pq terms is \begin{align} \frac{1}{2}(pq+1) \; where \; p ≠ q \end{align}
Ans:	It is known that the general term of an A.P. is an = a + (n – 1)d ∴ According to the given information, \begin{align}p^{th} \; term= a_p=a+(p-1)d=\frac{1}{q} \;\; ...(1)\end{align} \begin{align}q^{th} \; term= a_q=a+(q-1)d=\frac{1}{p} \;\; ...(2)\end{align} Subtracting (2) from (1), we obtain \begin{align} (p-1)d - (q-1)d=\frac{1}{q}-\frac{1}{p} \end{align} \begin{align} ⇒(p-1-q+1)d = \frac{p-q}{pq} \end{align} \begin{align} ⇒(p-q)d = \frac{p-q}{pq} \end{align} \begin{align} ⇒d = \frac{1}{pq} \end{align} Putting the value of d in (1), we obtain \begin{align} a + (p-1)\frac{1}{pq}= \frac{1}{q}\end{align} \begin{align} ⇒a = \frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq} \end{align} \begin{align} \therefore S_{pq} = \frac{pq}{2}\left[2a + (pq-1)d\right]\end{align} \begin{align} = \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\frac{1}{pq}\right]\end{align} \begin{align} = 1 + \frac{1}{2}(pq-1)\end{align} \begin{align} =\frac{1}{2}pq + 1 - \frac{1}{2}=\frac{1}{2}pq + \frac{1}{2}\end{align} \begin{align} =\frac{1}{2}(pq+1) \end{align} Thus, the sum of first pq terms of the A.P. is \begin{align} =\frac{1}{2}(pq+1). \end{align}

Q6	If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Ans:	Let the sum of n terms of the given A.P. be 116. \begin{align} S_n=\frac{n}{2}\left[2a + (n-1)d\right] \end{align} Here, a = 25 and d = 22 – 25 = – 3 \begin{align} \therefore S_n=\frac{n}{2}\left[2 × 25 + (n-1)(-3)\right] \end{align} \begin{align} ⇒ 116=\frac{n}{2}\left[50 -3n +3\right] \end{align} \begin{align} ⇒ 232=n(53-3n)=53n -3n^2 \end{align} \begin{align} ⇒ 3n^2 -53n + 232 =0\end{align} \begin{align} ⇒ 3n^2 -24n -29n + 232 =0\end{align} \begin{align} ⇒ 3n(n-8) -29(n-8) =0\end{align} \begin{align} ⇒ (n-8)(3n-29) =0\end{align} \begin{align} ⇒ n=8 \;or\;n=\frac{29}{3} \end{align} However, n cannot be equal to \begin{align} \frac{29}{3}. \end{align} Therefore, n = 8 \begin{align} \therefore a_8 = Last \; Term = a + (n-1)d= 25 + (8-1)(-3) \end{align} \begin{align} = 25 + (7)(-3)=25-21 \end{align} \begin{align} = 4 \end{align} Thus, the last term of the A.P. is 4.

Q7	Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Ans:	It is given that the kth term of the A.P. is 5k + 1. kth term = a_k = a + (k – 1)d ∴ a + (k – 1)d = 5k + 1 a + kd – d = 5k + 1 Comparing the coefficient of k, we obtain d = 5 a – d = 1 ⇒ a – 5 = 1 ⇒ a = 6 \begin{align} S_n = \frac{n}{2}\left[2a + (n-1)d\right] \end{align} \begin{align} = \frac{n}{2}\left[2(6) + (n-1)(5)\right] \end{align} \begin{align} = \frac{n}{2}\left[12 + 5n -5\right] \end{align} \begin{align} = \frac{n}{2}\left(5n + 7\right) \end{align}

Q8	If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference
Ans:	It is known that, \begin{align} S_n= \frac{n}{2}\left[2a + (n-1)d\right] \end{align} According to the given condition, \begin{align} \frac{n}{2}\left[2a + (n-1)d\right] = pn + qn^2 \end{align} \begin{align} ⇒ \frac{n}{2}\left[2a + nd-d\right] = pn + qn^2 \end{align} \begin{align} ⇒ na + n^2\frac{d}{2} - n.\frac{d}{2}= pn + qn^2 \end{align} Comparing the coefficients of n² on both sides, we obtain \begin{align} \frac{d}{2} = q \end{align} \begin{align} \therefore d = 2q \end{align} Thus, the common difference of the A.P. is 2q.

Q9	The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
Ans:	Let a₁, a₂, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively. According to the given condition, \begin{align} \frac{Sum \;of \;n \;terms \;of \;first\; A.P.}{Sum\; of \;n\; terms \;of \;second \;A.P.} = \frac{5n+4}{9n+6} \end{align} \begin{align} ⇒\frac{\frac{n}{2}\left[2a_1 + (n-1)d_1\right]}{\frac{n}{2}\left[2a_2 + (n-1)d_2\right]} = \frac{5n+4}{9n+6} \end{align} \begin{align} ⇒\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6} \;\;\;\;...(1)\end{align} Substituting n = 35 in (1), we obtain \begin{align} ⇒\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6} \end{align} \begin{align} ⇒\frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321} \;\;\;\;...(2)\end{align} \begin{align} \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{a_1 + 17d_1}{a_2 + 17d_2} \;\;\;\;...(3)\end{align} From (2) and (3), we obtain \begin{align} \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{179}{321}\end{align} Thus, the ratio of 18th term of both the A.P.s is 179: 321.

Q10	If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Ans:	Let a and d be the first term and the common difference of the A.P. respectively. Here, \begin{align} S_p = \frac{p}{2}\left[2a+(p-1)d\right]\end{align} \begin{align} S_q = \frac{q}{2}\left[2a+(q-1)d\right]\end{align} According to the given condition, \begin{align} \frac{p}{2}\left[2a+(p-1)d\right]=\frac{q}{2}\left[2a+(q-1)d\right]\end{align} \begin{align} ⇒p\left[2a+(p-1)d\right]=q\left[2a+(q-1)d\right]\end{align} \begin{align} ⇒2ap + pd(p-1)=2aq+qd(q-1)\end{align} \begin{align} ⇒2a(p-q) +d[p(p-1)-q(q-1)]=0\end{align} \begin{align} ⇒2a(p-q) +d[p^2 -p -q^2 +q]=0\end{align} \begin{align} ⇒2a(p-q) +d[(p-q)(p+q)-(p-q)]=0\end{align} \begin{align} ⇒2a(p-q) +d[(p-q)(p+q-1)]=0\end{align} \begin{align} ⇒2a +d(p+q-1)=0\end{align} \begin{align} ⇒d=\frac{-2a}{p+q-1} \;\;\;\;...(1)\end{align} \begin{align} \therefore S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).d\right] \end{align} \begin{align} ⇒ S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).\left(\frac{-2a}{p+q-1}\right)\right] \;\;\;\; [From (1)] \end{align} \begin{align}=\frac{p+q}{2}\left[2a-2a\right]\end{align} \begin{align}=0\end{align} Thus, the sum of the first (p + q) terms of the A.P. is 0.

Q11	Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that $a over p open parentheses q minus r close parentheses plus b over q open parentheses r minus p close parentheses plus c over r open parentheses p minus q close parentheses equals 0$
Ans:	Let a₁ and d be the first term and the common difference of the A.P. respectively. According to the given information, $S subscript p space end subscript equals space p over 2 open square brackets 2 a subscript 1 plus space open parentheses p minus 1 close parentheses d close square brackets equals a rightwards double arrow 2 a subscript 1 space plus space open parentheses p minus 1 close parentheses d space equals space fraction numerator 2 a over denominator p end fraction space space space space space space... left parenthesis 1 right parenthesis S subscript q equals space q over 2 open square brackets 2 a subscript 1 plus space open parentheses q minus 1 close parentheses d close square brackets equals b rightwards double arrow 2 a subscript 1 space plus space open parentheses q minus 1 close parentheses d space equals space fraction numerator 2 b over denominator q end fraction space space space space space space... left parenthesis 2 right parenthesis S subscript r equals space r over 2 open square brackets 2 a subscript 1 plus space open parentheses r minus 1 close parentheses d close square brackets equals c rightwards double arrow 2 a subscript 1 space plus space open parentheses r minus 1 close parentheses d space equals space fraction numerator 2 c over denominator r end fraction space space space space space space... left parenthesis 3 right parenthesis S u b t r a c t i n g space left parenthesis 2 right parenthesis space f r o m space left parenthesis 1 right parenthesis comma space w e space o b t a i n left parenthesis p minus 1 right parenthesis d space minus left parenthesis q minus 1 right parenthesis d equals fraction numerator 2 a over denominator p end fraction equals fraction numerator 2 b over denominator q end fraction rightwards double arrow d open parentheses p minus 1 minus q plus 1 close parentheses equals fraction numerator 2 a q minus 2 b q over denominator p q end fraction rightwards double arrow d left parenthesis p minus q right parenthesis equals fraction numerator 2 a q minus 2 b p over denominator p q end fraction rightwards double arrow d equals fraction numerator 2 open parentheses a q minus b p close parentheses over denominator p q open parentheses p minus q close parentheses end fraction space space space space space space space space space space space... left parenthesis 4 right parenthesis S u b t r a c t i n g space left parenthesis 3 right parenthesis space f r o m space left parenthesis 2 right parenthesis comma space w e space o b t a i n open parentheses q minus 1 close parentheses d minus open parentheses r minus 1 close parentheses d equals fraction numerator 2 b over denominator q end fraction minus fraction numerator 2 c over denominator r end fraction rightwards double arrow d open parentheses q minus 1 minus r plus 1 close parentheses equals fraction numerator 2 b over denominator q end fraction minus fraction numerator 2 c over denominator r end fraction rightwards double arrow d open parentheses q minus r close parentheses equals fraction numerator 2 b r minus 2 q c over denominator q r end fraction rightwards double arrow d equals fraction numerator 2 open parentheses b r minus q c close parentheses over denominator q r open parentheses q minus r close parentheses end fraction space space space space space space space space space space space... left parenthesis 5 right parenthesis E q u a t i n g space b o t h space t h e space v a l u e space o f space d space o b t a i n e d space i n space left parenthesis 4 right parenthesis space a n d space left parenthesis 5 right parenthesis comma space w e space o b t a i n fraction numerator a q minus b p over denominator p q open parentheses p minus q close parentheses end fraction equals fraction numerator b r minus q c over denominator q r open parentheses q minus r close parentheses end fraction rightwards double arrow q r open parentheses q minus r close parentheses open parentheses a q minus b q close parentheses equals p q open parentheses p minus q close parentheses open parentheses b r minus q c close parentheses rightwards double arrow r open parentheses a q minus b p close parentheses open parentheses q minus r close parentheses equals p open parentheses b r minus q c close parentheses open parentheses p minus q close parentheses rightwards double arrow open parentheses a q r minus b p r close parentheses open parentheses q minus r close parentheses equals open parentheses b p r minus p q c close parentheses open parentheses p minus q close parentheses D i v i d i n g space b o t h space s i d e s space b space p q r ; w e space o b a t i n open parentheses a over p minus b over q close parentheses open parentheses q minus r close parentheses equals open parentheses b over q minus c over r close parentheses open parentheses p minus q close parentheses rightwards double arrow a over p open parentheses q minus r close parentheses minus b over q open parentheses q minus r plus p minus q close parentheses plus c over r open parentheses p minus q close parentheses equals 0 rightwards double arrow a over p open parentheses q minus r close parentheses plus b over q open parentheses r minus p close parentheses plus c over r open parentheses p minus q close parentheses equals 0 T h u s comma t h e space g i v e n space r e s u l t space i s space p r o v e d.$

Q12	The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).
Ans:	Let a and b be the first term and the common difference of the A.P. respectively. According to the given condition, $fraction numerator S u m space o f space m space t e r m s over denominator s u m space o f space n space t e r m s end fraction equals m squared over n squared rightwards double arrow fraction numerator begin display style m over 2 end style open square brackets 2 a plus open parentheses m minus 1 close parentheses d close square brackets over denominator begin display style n over 2 end style open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets end fraction equals m squared over n squared rightwards double arrow space space space fraction numerator 2 a plus left parenthesis m minus 1 right parenthesis d over denominator 2 a plus left parenthesis n minus 1 right parenthesis d end fraction equals m over n space space space space space.... left parenthesis 1 right parenthesis P u t t i n g space m space equals space 2 m space � space 1 space a n d space n space equals space 2 n space � space 1 space i n space left parenthesis 1 right parenthesis comma space w e space o b t a i n fraction numerator 2 a plus left parenthesis 2 m minus 2 right parenthesis d over denominator 2 a plus left parenthesis 2 n minus 2 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction rightwards double arrow fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction space space space space.... left parenthesis 2 right parenthesis fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction space space space space... left parenthesis 3 right parenthesis F r o m space left parenthesis 2 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space o b t a i n fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction T h u s comma space t h e space g i v e n space r e s u l t space i s space p r o v e d.$

Q13	If the sum of n terms of an A.P. is 3n² + 5n and its mth term is 164, find the value of m.
Ans:	Let a and b be the first term and the common difference of the A.P. respectively. am = a + (m – 1)d = 164 … (1) Sum of n terms, $S subscript n space space end subscript equals space n over 2 open square brackets 2 a plus left parenthesis n minus 1 right parenthesis d close square brackets$ Here, $n over 2 open square brackets 2 a plus n d minus d close square brackets equals 3 n squared plus 5 n rightwards double arrow n a space plus space n squared. d over 2 equals 3 n squared plus 5 n$ Comparing the coefficient of n2 on both sides, we obtain $d over 2 equals 3 rightwards double arrow d equals 6$ Comparing the coefficient of n on both sides, we obtain $a minus d over 2 equals 5 rightwards double arrow a minus 3 equals 5 rightwards double arrow a equals 8$ Therefore, from (1), we obtain 8 + (m – 1) 6 = 164 ⇒ (m – 1) 6 = 164 – 8 = 156 ⇒ m – 1 = 26 ⇒ m = 27 Thus, the value of m is 27.

Q14	Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Ans:	Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P. Here, a = 8, b = 26, n = 7 Therefore, 26 = 8 + (7 – 1) d ⇒ 6d = 26 – 8 = 18 ⇒ d = 3 A1 = a + d = 8 + 3 = 11 A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14 A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17 A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20 A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23 Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Q15	If $fraction numerator a to the power of n plus b to the power of n over denominator a to the power of n minus 1 end exponent plus b to the power of n minus 1 end exponent end fraction$ is the A.M. between a and b, then find the value of n.
Ans:	A.M. of a and b = $fraction numerator a plus b over denominator 2 end fraction$ According to the given condition, $fraction numerator a plus b over denominator 2 end fraction equals fraction numerator a to the power of n space space space end exponent plus space b to the power of n over denominator a to the power of n minus 1 end exponent plus b to the power of n minus 1 end exponent end fraction rightwards double arrow open parentheses a plus b close parentheses open parentheses a to the power of n minus 1 end exponent plus b to the power of n minus 1 end exponent close parentheses equals 2 open parentheses a to the power of n plus b to the power of n close parentheses rightwards double arrow a to the power of n plus a b to the power of n minus 1 end exponent plus b a to the power of n minus 1 end exponent plus b to the power of n equals 2 a to the power of n plus 2 b to the power of n rightwards double arrow a b to the power of n minus 1 end exponent plus a to the power of n minus 1 end exponent b equals a to the power of n plus b to the power of n rightwards double arrow a b to the power of n minus 1 end exponent minus b to the power of n equals a to the power of n minus a to the power of n minus 1 end exponent b rightwards double arrow b to the power of n minus 1 end exponent open parentheses a minus b close parentheses equals a to the power of n minus 1 end exponent open parentheses a minus b close parentheses rightwards double arrow b to the power of n minus 1 end exponent equals a to the power of n minus 1 end exponent rightwards double arrow open parentheses a over b close parentheses to the power of n minus 1 end exponent equals 1 equals open parentheses a over b close parentheses to the power of 0 rightwards double arrow n minus 1 equals 0 rightwards double arrow n equals 1$

Q16	Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)thnumbers is 5:9. Find the value of m.
Ans:	Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P. Here, a = 1, b = 31, n = m + 2 ∴ 31 = 1 + (m + 2 – 1) (d) ⇒ 30 = (m + 1) d $rightwards double arrow d equals fraction numerator 30 over denominator m plus 1 end fraction space space space space space space space space... left parenthesis 1 right parenthesis$ A1 = a + d A2 = a + 2d A3 = a + 3d … ∴ A7 = a + 7d Am–1 = a + (m – 1) d According to the given condition, $fraction numerator a plus 7 d over denominator a plus open parentheses m minus 1 close parentheses d end fraction equals 5 over 9 rightwards double arrow fraction numerator 1 plus 7 open parentheses begin display style fraction numerator 30 over denominator open parentheses m plus 1 close parentheses end fraction end style close parentheses over denominator 1 plus open parentheses m minus 1 close parentheses open parentheses begin display style fraction numerator 30 over denominator m plus 1 end fraction end style close parentheses end fraction equals 5 over 9 space space space space space space space space space space space space space space space space open square brackets F r o m space left parenthesis 1 right parenthesis close square brackets rightwards double arrow fraction numerator m plus 1 plus 7 open parentheses 30 close parentheses over denominator m plus 1 plus 30 open parentheses m minus 1 close parentheses end fraction equals 5 over 9 rightwards double arrow fraction numerator m plus 1 plus 210 over denominator m plus 1 plus 30 m minus 30 end fraction equals 5 over 9 rightwards double arrow fraction numerator m plus 211 over denominator 31 m space minus 29 end fraction equals 5 over 9 rightwards double arrow 9 m space plus space 1899 space equals space 155 m space minus 145 rightwards double arrow 155 m space minus 9 m space equals space 1899 space plus 145 rightwards double arrow 146 m equals 2044 rightwards double arrow m equals 14$ Thus, the value of m is 14.

Q17	A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?
Ans:	The first installment of the loan is Rs 100. The second installment of the loan is Rs 105 and so on. The amount that the man repays every month forms an A.P. The A.P. is 100, 105, 110, … First term, a = 100 Common difference, d = 5 A30 = a + (30 – 1)d = 100 + (29) (5) = 100 + 145 = 245 Thus, the amount to be paid in the 30th installment is Rs 245.

Q18	The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Ans:	The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°. It is known that the sum of all angles of a polygon with n sides is 180° (n – 2). So $S subscript n equals 180 to the power of 0 open parentheses n minus 2 close parentheses rightwards double arrow n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets equals 180 to the power of 0 open parentheses n minus 2 close parentheses rightwards double arrow n over 2 open square brackets 240 to the power of 0 plus open parentheses n minus 1 close parentheses 5 to the power of 0 close square brackets equals 180 open parentheses n minus 2 close parentheses rightwards double arrow n open square brackets 240 plus open parentheses n minus 1 close parentheses 5 close square brackets equals 360 open parentheses n minus 2 close parentheses rightwards double arrow 240 n space plus space 5 n squared space minus 5 n space equals space 360 n space minus 720 rightwards double arrow 5 n squared plus 235 n minus 360 n space plus 720 equals 0 rightwards double arrow 5 n squared space minus 125 n space plus 720 equals 0 rightwards double arrow n squared minus 25 n space plus space 144 space equals space 0 rightwards double arrow n squared minus 16 n minus 9 n plus 144 equals 0 rightwards double arrow n open parentheses n minus 16 close parentheses minus 9 open parentheses n minus 16 close parentheses equals 0 rightwards double arrow open parentheses n minus 9 close parentheses open parentheses n minus 16 close parentheses equals 0 rightwards double arrow n equals 9 space o r space 16$

Exercise 3

Q1	Find the 20th and nthterms of the G.P. $5 over 2 comma 5 over 4 comma 5 over 8 comma...$
Ans:	Here, a = First Term = $5 over 2$ r = common ratio = $fraction numerator begin display style 5 over 4 end style over denominator begin display style 5 over 2 end style end fraction equals 1 half$ $a subscript 20 space space end subscript equals space a r to the power of 20 minus 1 end exponent equals 5 over 2 open parentheses 1 half close parentheses to the power of 19 equals fraction numerator 5 over denominator open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of 19 end fraction equals 5 over open parentheses 2 close parentheses to the power of 20 a subscript n space equals space a r to the power of n minus 1 end exponent equals 5 over 2 open parentheses 1 half close parentheses to the power of n minus 1 end exponent equals fraction numerator 5 over denominator open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of n minus 1 end exponent end fraction equals 5 over open parentheses 2 close parentheses to the power of n$

Q2	Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Ans:	Common ratio, r = 2 Let a be the first term of the G.P. ∴ a⁸ = ar ^8–1 = ar⁷ ⇒ ar⁷ = 192 a(2)⁷ = 192 a(2)⁷ = (2)⁶ (3) $rightwards double arrow a equals fraction numerator open parentheses 2 close parentheses to the power of 6 cross times 3 space over denominator open parentheses 2 close parentheses to the power of 7 end fraction equals 3 over 2 s o a subscript 12 space equals a r to the power of 12 minus 1 end exponent space equals space open parentheses 3 over 2 close parentheses open parentheses 2 close parentheses to the power of 11 equals open parentheses 3 close parentheses open parentheses 2 close parentheses to the power of 10 equals 3072$

Q3	The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Ans:	Let a be the first term and r be the common ratio of the G.P. According to the given condition, a₅ = a r⁵–1 = a r⁴ = p … (1) a₈ = a r⁸–1 = a r⁷ = q … (2) a₁₁ = a r¹¹–1 = a r¹⁰ = s … (3) Dividing equation (2) by (1), we obtain $fraction numerator a r to the power of 7 over denominator a r to the power of 4 end fraction equals q over p r cubed equals q over p space space space space space space... left parenthesis 4 right parenthesis$ Dividing equation (3) by (2), we obtain $fraction numerator a r to the power of 10 over denominator a r to the power of 7 end fraction equals s over q rightwards double arrow r cubed equals s over q space space space space space space space... left parenthesis 5 right parenthesis$ Equating the values of r3 obtained in (4) and (5), we obtain $q over p equals s over q rightwards double arrow q squared equals p s$ Thus, the given result is proved.

Q4	The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Ans:	Let a be the first term and r be the common ratio of the G.P. ∴ a = –3 It is known that, an = arn–1 ∴a₄ = ar³ = (–3) r³ a₂ = a r¹ = (–3) r According to the given condition, (–3) r³ = [(–3) r]² ⇒ –3r³ = 9 r² ⇒ r = –3 a₇ = a r ⁷–1 = a r⁶ = (–3) (–3)6 = – (3)7 = –2187 Thus, the seventh term of the G.P. is –2187.

Q5	Which term of the following sequences: $2 comma space 2 square root of 2 comma 4 comma space...$
Ans:	(a) The given sequence is $2 comma space 2 square root of 2 comma 4 comma space...$ Here, a = 2 and r = $fraction numerator 2 square root of 2 over denominator 2 end fraction equals square root of 2$ Let the nth term of the given sequence be 128. $a subscript n space end subscript equals space a r to the power of n minus 1 end exponent rightwards double arrow open parentheses 2 close parentheses open parentheses square root of 2 close parentheses to the power of n minus 1 end exponent space equals space 128 rightwards double arrow open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 2 close parentheses to the power of 7 rightwards double arrow open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction plus 1 end exponent equals open parentheses 2 close parentheses to the power of 7 s o space fraction numerator n minus 1 over denominator 2 end fraction plus 1 equals 7 rightwards double arrow fraction numerator n minus 1 over denominator 2 end fraction space equals space 6 rightwards double arrow n space minus 1 space equals space 12 rightwards double arrow n equals 13$ Thus, the 13th term of the given sequence is 128. (b) The given sequence is $square root of 3 space comma space 3 comma space 3 square root of 3 space comma space... space i s space 729 ?$ Here, a= $square root of 3 space a n d space r space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3$ Let the nth term of the given sequence be 729. $a subscript n equals a r to the power of n minus 1 end exponent s o a r to the power of n minus 1 end exponent space equals space 729 rightwards double arrow open parentheses square root of 3 close parentheses space open parentheses square root of 3 close parentheses to the power of n minus 1 end exponent space equals space 729 rightwards double arrow open parentheses 3 close parentheses to the power of 1 half end exponent open parentheses 3 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6 rightwards double arrow open parentheses 3 close parentheses to the power of 1 half plus fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6 s o 1 half plus fraction numerator n minus 1 over denominator 2 end fraction equals 6 rightwards double arrow fraction numerator 1 plus n minus 1 over denominator 2 end fraction equals 6 rightwards double arrow n equals 12$ Thus, the 12th term of the given sequence is 729. (c) The given sequence is $1 third comma space 1 over 9 comma space 1 over 27 comma space... space$ Here, a= $1 third space space a n d space r space equals space 1 over 9 obelus divided by 1 third equals 1 third$ Let the nth term of the given sequence be $1 over 19683$ . $a subscript n space equals space a r to the power of n minus 1 end exponent s o a r to the power of n minus 1 end exponent equals 1 over 19683 rightwards double arrow open parentheses 1 third close parentheses open parentheses 1 third close parentheses to the power of n minus 1 end exponent equals 1 over 19683 rightwards double arrow open parentheses 1 third close parentheses to the power of n equals open parentheses 1 third close parentheses to the power of 9 rightwards double arrow n equals 9$ Thus, the 9th term of the given sequence is $1 over 19683$ .

Q6	For what values of x, the numbers $2 over 7 comma x comma fraction numerator minus 7 over denominator 2 end fraction space$ are in G.P?
Ans:	The given numbers are $fraction numerator minus 2 over denominator 7 end fraction comma space x comma space fraction numerator minus 7 over denominator 2 end fraction$ . Common ratio = $fraction numerator x over denominator begin display style fraction numerator minus 2 over denominator 7 end fraction end style end fraction equals fraction numerator minus 7 x over denominator 2 end fraction$ Also, common ratio = $fraction numerator minus 7 over denominator begin display style 2 over x end style end fraction equals fraction numerator minus 7 over denominator 2 x end fraction$ so $fraction numerator minus 7 x over denominator 2 end fraction equals fraction numerator minus 7 over denominator 2 x end fraction rightwards double arrow x squared equals fraction numerator minus 2 space cross times 7 over denominator minus 2 space cross times 7 end fraction space equals 1 rightwards double arrow x space equals space square root of 1 space rightwards double arrow x space equals space plus-or-minus 1 space$ Thus, for x = ± 1, the given numbers will be in G.P.

Q7	Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Ans:	The given G.P. is 0.15, 0.015, 0.00015, … Here, a = 0.15 and r = $fraction numerator 0.015 over denominator 0.15 end fraction equals 0.1$ $S subscript n space end subscript equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction s o S subscript 20 space equals space fraction numerator 0.15 open square brackets 1 minus open parentheses 0.1 close parentheses to the power of 20 close square brackets over denominator 1 minus 0.1 end fraction space space space space space space space space space equals space fraction numerator 0.15 over denominator 0.9 end fraction open square brackets 1 minus open parentheses 0.1 close parentheses to the power of 20 close square brackets space space space space space space space space space equals space 15 over 90 space open square brackets 1 space minus space open parentheses 0.1 close parentheses to the power of 20 close square brackets space space space space space space space space space equals space 1 over 6 open square brackets 1 minus open parentheses 0.1 to the power of 20 close parentheses close square brackets space$

Q8	Find the sum to n terms in the geometric progression $square root of 7 space comma space square root of 21 space comma space 3 square root of 7 space comma space...$
Ans:	The given G.P. is $square root of 7 space comma space square root of 21 space comma space 3 square root of 7 space comma space...$ Here, a = $square root of 7$ $r space equals fraction numerator square root of 21 over denominator square root of 7 end fraction space equals space square root of 3 S subscript n space space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction S o S subscript n space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction space space space space space space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction cross times fraction numerator 1 plus square root of 3 over denominator 1 space plus space square root of 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses B y space r a t i o n a l i z i n g close parentheses space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 space plus square root of 3 close parentheses open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus 3 end fraction space space space space space space space space equals space fraction numerator minus square root of 7 space open parentheses 1 space plus space square root of 3 close parentheses over denominator 2 end fraction open square brackets 1 minus open parentheses 3 close parentheses to the power of n over 2 end exponent close square brackets space space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 plus square root of 3 close parentheses over denominator 2 end fraction open square brackets open parentheses 3 close parentheses to the power of n over 2 end exponent minus 1 close square brackets$

Q9	Find the sum to n terms in the geometric progression 1,-a, a²,-a³, ... (if a ≠ -1)
Ans:	The given G.P. is 1,-a, a²,-a³, ......... Here, first term = a1 = 1 Common ratio = r = – a $S subscript n equals fraction numerator a subscript 1 open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction s o S subscript n equals fraction numerator 1 open square brackets 1 minus open parentheses minus a close parentheses to the power of n close square brackets over denominator 1 minus open parentheses minus a close parentheses end fraction space equals space fraction numerator open square brackets 1 minus open parentheses minus a close parentheses to the power of n close square brackets over denominator 1 plus a end fraction$

Q10	Find the sum to n terms in the geometric progression x³, x⁵, x⁷ ... (if x ≠ ±1)
Ans:	The given G.P. is x³, x⁵, x⁷ ........ Here, a = x³ and r = x² ^{$S subscript n space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction equals fraction numerator x cubed open square brackets 1 minus open parentheses x squared close parentheses to the power of n close square brackets over denominator 1 minus x squared end fraction equals fraction numerator x cubed open parentheses 1 minus x to the power of 2 n end exponent close parentheses over denominator 1 minus x squared end fraction$}

Q11	Evaluate $stack sum open parentheses 2 plus 3 to the power of k close parentheses with k equals 1 below and 11 on top$
Ans:	$stack sum space with k equals 1 below and 11 on top open parentheses 2 plus 3 to the power of k close parentheses equals sum from k equals 1 to 11 of open parentheses 2 close parentheses space plus space sum from k equals 1 to 11 of 3 to the power of k space equals space 2 open parentheses 11 close parentheses space plus space sum from k equals 1 to 11 of 3 to the power of k to the power of blank end exponent space equals 22 space plus space sum from k equals 1 to 11 of 3 to the power of k space space space space space space... open parentheses 1 close parentheses sum from k equals 1 to 11 of 3 to the power of k space equals space 3 to the power of 1 space plus space 3 squared space plus space 3 cubed space plus space... space plus space 3 to the power of 11 T h e space t e r m s space o f space t h i s space s e q u e n c e space 3 comma space 3 squared comma space 3 cubed comma space horizontal ellipsis space f o r m s space a space G. P. S subscript n space end subscript equals space fraction numerator a open parentheses r to the power of n space end exponent minus 1 close parentheses over denominator r minus 1 end fraction rightwards double arrow S subscript n space equals space fraction numerator 3 open square brackets open parentheses 3 close parentheses to the power of 11 space minus 1 close square brackets over denominator 3 minus 1 end fraction rightwards double arrow S subscript n space equals space 3 over 2 open parentheses 3 to the power of 11 space minus space 1 close parentheses therefore space space sum from k equals 1 to 11 of 3 to the power of k space equals space 3 over 2 open parentheses 3 to the power of 11 space minus 1 close parentheses S u b s t i t u t i n g space t h i s space v a l u e space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space o b t a i n sum from k equals 1 to 11 of open parentheses 2 space plus space 3 to the power of k close parentheses space equals 22 space plus space 3 over 2 open parentheses 3 to the power of 11 space minus 1 close parentheses$

Q12	The sum of first three terms of a G.P. is $39 over 10$ and their product is 1. Find the common ratio and the terms.
Ans:	Let $a over r space comma space a comma space a r$ be the first three terms of the G.P. $a over r space plus space a space plus space a r space equals space 39 over 10 space space space space space space... space left parenthesis 1 right parenthesis open parentheses a over r close parentheses open parentheses a close parentheses open parentheses a r close parentheses space equals space 1 space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis$ From (2), we obtain a³ = 1 ⇒ a = 1 (Considering real roots only) Substituting a = 1 in equation (1), we obtain $1 over r space plus space 1 space plus space r space equals space 39 over 10 rightwards double arrow 1 space plus space r space plus space r squared space equals space 39 over 10 r rightwards double arrow 10 space plus space 10 r space plus space 10 r squared space minus 39 r space equals 0 rightwards double arrow 10 r squared space minus 29 r space plus space 10 space equals 0 rightwards double arrow space 10 r squared space minus space 25 r space minus space 4 r space plus space 10 space equals space 0 rightwards double arrow 5 r space open parentheses 2 r space minus 5 close parentheses space minus space 2 open parentheses 2 r space minus space 5 close parentheses space equals space 0 rightwards double arrow open parentheses 5 r space minus space 2 close parentheses open parentheses 2 r minus 5 close parentheses space equals space 0 rightwards double arrow r space equals space 2 over 5 space o r space 5 over 2 T h u s comma space t h e space t h r e e space t e r m s space o f space G. P. space a r e space 5 over 2 comma space 1 space comma space a n d space 2 over 5. space space$

Q13	How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?
Ans:	The given G.P. is 3, 3², 3³, … Let n terms of this G.P. be required to obtain the sum as 120. $S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r minus 1 end fraction$ Here, a = 3 and r = 3 ∴ $S subscript n space equals space 120 space equals space fraction numerator 3 space open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction rightwards double arrow 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 2 end fraction rightwards double arrow fraction numerator 120 space cross times 2 over denominator 3 end fraction space equals space 3 to the power of n space minus 1 rightwards double arrow 3 to the power of n space minus 1 space equals space 80 rightwards double arrow 3 to the power of n space equals space 81 rightwards double arrow 3 to the power of n space equals space 3 to the power of 4$ ∴ n = 4 Thus, four terms of the given G.P. are required to obtain the sum as 120.

Q14	The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Ans:	Let the G.P. be a, ar, ar², ar³, … According to the given condition, a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128 ⇒ a (1 + r + r2) = 16 … (1) ar³(1 + r + r2) = 128 … (2) Dividing equation (2) by (1), we obtain $fraction numerator a r cubed open parentheses 1 plus r plus r squared close parentheses over denominator a open parentheses 1 plus r plus r squared close parentheses end fraction space equals space 128 over 16 rightwards double arrow space r cubed space equals space 8 therefore space r space equals space 2$ Substituting r = 2 in (1), we obtain a (1 + 2 + 4) = 16 ⇒ a (7) = 16 $rightwards double arrow a space equals space 16 over 7 S subscript n space end subscript equals space fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction rightwards double arrow S subscript n space equals space fraction numerator 16 space over denominator 7 end fraction fraction numerator open parentheses 2 to the power of n space minus 1 close parentheses over denominator 2 minus 1 end fraction space equals space 16 over 7 space open parentheses 2 to the power of n space minus 1 close parentheses$

Q15	Given a G.P. with a = 729 and 7th term 64, determine S₇.
Ans:	a = 729 a₇ = 64 Let r be the common ratio of the G.P. It is known that, a_n = a r^n–1 a₇ = ar^7–1 = (729)r⁶ ⇒ 64 = 729 r⁶ ^{$rightwards double arrow r to the power of 6 space equals space 64 over 729 rightwards double arrow r to the power of 6 space space equals space open parentheses 2 over 3 close parentheses to the power of 6 rightwards double arrow r space equals space 2 over 3$} Also, it is known that, $S subscript n space equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction therefore space S subscript 7 space equals space fraction numerator 729 open square brackets 1 minus open parentheses begin display style 2 over 3 end style close parentheses to the power of 7 close square brackets over denominator 1 minus begin display style 2 over 3 end style end fraction space space space space space space space space space space space space equals space 3 space cross times space 729 space open square brackets 1 minus open parentheses 2 over 3 close parentheses to the power of 7 close square brackets space space space space space space space space space space space space equals open parentheses 3 close parentheses to the power of 7 space open square brackets fraction numerator open parentheses 3 close parentheses to the power of 7 space minus space open parentheses 2 close parentheses to the power of 7 over denominator open parentheses 3 close parentheses to the power of 7 end fraction close square brackets space space space space space space space space space space space space equals space open parentheses 3 close parentheses to the power of 7 space minus open parentheses 2 close parentheses to the power of 7 space space space space space space space space space space space space space equals space 2187 space minus space 128 space space space space space space space space space space space space space equals space 2059$

Q16	Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.
Ans:	Let a be the first term and r be the common ratio of the G.P. According to the given conditions, $S subscript 2 space equals space minus 4 space equals space fraction numerator a open parentheses 1 minus r squared close parentheses over denominator 1 minus r end fraction space space space space space space space space space... left parenthesis 1 right parenthesis a subscript 5 space equals space 4 space cross times space a subscript 3 a r to the power of 4 space equals space 4 a r squared rightwards double arrow r squared space equals space 4 therefore space r space equals space plus-or-minus 2$ From (1), we obtain $minus 4 space equals space fraction numerator a open square brackets 1 minus open parentheses 2 close parentheses squared close square brackets over denominator 1 minus 2 end fraction space f o r space r space equals 2 rightwards double arrow minus 4 space equals space fraction numerator a open parentheses 1 minus 4 close parentheses over denominator minus 1 end fraction rightwards double arrow minus 4 space equals space a open parentheses 3 close parentheses rightwards double arrow a space equals space fraction numerator minus 4 over denominator 3 end fraction A l s o comma space minus 4 space equals space fraction numerator a open square brackets 1 minus open parentheses minus 2 close parentheses squared close square brackets over denominator 1 minus open parentheses minus 2 close parentheses end fraction space f o r space r space equals minus 2 rightwards double arrow minus 4 space equals space fraction numerator a open parentheses 1 minus 4 close parentheses over denominator 1 plus 2 end fraction rightwards double arrow minus 4 space equals space fraction numerator a open parentheses minus 3 close parentheses over denominator 3 end fraction rightwards double arrow a space equals space 4 T h u s comma space t h e space r e q u i r e d space G. P. space i s fraction numerator minus 4 over denominator 3 end fraction comma space fraction numerator minus 8 over denominator 3 end fraction comma space fraction numerator minus 16 over denominator 3 end fraction space comma space... space o r space 4 comma space minus 8 comma space 16 comma space minus 32 comma space...$ .....

Q17	If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Ans:	Let a be the first term and r be the common ratio of the G.P. According to the given condition, a₄ = a r³ = x … (1) a₁₀ = a r⁹ = y … (2) a₁₆ = a r¹⁵ = z … (3) Dividing (2) by (1), we obtain $y over x space equals space fraction numerator a r to the power of 9 over denominator a r cubed end fraction space rightwards double arrow y over x space equals space r to the power of 6$ Dividing (3) by (2), we obtain $z over y space equals space fraction numerator a r to the power of 15 over denominator a r to the power of 9 end fraction rightwards double arrow z over y equals r to the power of 6$ ∴ $y over x space equals space z over y$ Thus, x, y, z are in G. P.

Q18	Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Ans:	The given sequence is 8, 88, 888, 8888… This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as S_n = 8 + 88 + 888 + 8888 + …………….. to n terms $equals 8 over 9 open square brackets 9 plus 99 plus 999 plus 9999 plus.......... t o space n space t e r m s close square brackets equals 8 over 9 open square brackets open parentheses 10 minus 1 close parentheses plus open parentheses 10 squared minus 1 close parentheses plus open parentheses 10 cubed minus 1 close parentheses plus open parentheses 10 to the power of 4 minus 1 close parentheses plus........ t o to the power of space n space t e r m s close square brackets equals 8 over 9 open square brackets open parentheses 10 plus 10 squared space plus space....... n space t e r m s close parentheses space minus space open parentheses 1 plus 1 plus 1 plus... n space t e r m s close parentheses close square brackets equals 8 over 9 open square brackets fraction numerator 10 open parentheses 10 to the power of n minus 1 close parentheses over denominator 10 minus 1 end fraction space minus space n close square brackets equals 8 over 9 open square brackets fraction numerator 10 open parentheses 10 to the power of n minus 1 close parentheses over denominator 9 end fraction space minus space n close square brackets equals 80 over 81 open parentheses 10 to the power of n minus 1 close parentheses space minus space 8 over 9 n$

Q19	Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, $1 half$ .
Ans:	Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × $1 half$ $equals space 64 space open square brackets 4 space plus space 2 space plus space 1 space plus 1 half space plus space 1 over 2 squared close square brackets$ Here, 4, 2, 1, $1 half comma space 1 over 2 squared$ is a G.P. First term, a = 4 Common ratio, r = $1 half$ It is known that, $S subscript n space equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction therefore space S subscript 5 space equals space fraction numerator 4 open square brackets 1 minus open parentheses begin display style 1 half end style close parentheses to the power of 5 close square brackets over denominator 1 minus begin display style 1 half end style end fraction equals fraction numerator 4 open square brackets 1 minus begin display style 1 over 32 end style close square brackets over denominator begin display style 1 half end style end fraction equals fraction numerator 8 open parentheses 32 minus 1 close parentheses over denominator 32 end fraction equals 31 over 4 therefore space R e q u i r e d space s u m space equals space 64 space open parentheses 31 over 4 close parentheses space equals space open parentheses 16 close parentheses open parentheses 31 close parentheses space equals 496$

Q20	Show that the products of the corresponding terms of the sequences a,ar,ar², ...ar^n-1 and A, AR, AR², ,,,AR^n-1form a G.P, and find the common ratio.
Ans:	It has to be proved that the sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P. $fraction numerator S e c o n d space t e r m over denominator F i r s t space t e r m end fraction space equals space fraction numerator a r A R over denominator a A end fraction equals r R fraction numerator T h i r d space t e r m over denominator S e c o n d space t e r m end fraction space equals space fraction numerator a r squared A R squared over denominator a r A R end fraction equals r R$ Thus, the above sequence forms a G.P. and the common ratio is rR.

Q21	Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Ans:	Let a be the first term and r be the common ratio of the G.P. a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³ By the given condition, a₃ = a₁ + 9 ⇒ ar² = a + 9 … (1) a₂ = a₄ + 18 ⇒ ar = ar³ + 18 … (2) From (1) and (2), we obtain a(r² – 1) = 9 … (3) ar (1– r²) = 18 … (4) Dividing (4) by (3), we obtain $fraction numerator a r open parentheses 1 minus r squared close parentheses over denominator a open parentheses r squared minus 1 close parentheses end fraction equals 18 over 9 rightwards double arrow minus r space equals space 2 rightwards double arrow r space equals space minus 2$ Substituting the value of r in (1), we obtain 4a = a + 9 ⇒ 3a = 9 ∴ a = 3 Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)², and 3(–2)³ i.e., 3¸–6, 12, and –24.

Q22	If the p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^q-rb^r-pc^p-q=1
Ans:	Let A be the first term and R be the common ratio of the G.P. According to the given information, ARp–1 = a ARq–1 = b ARr–1 = c aq–r br–p cp–q = Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r-p) × Ap–q × R(r –1)(p–q) = Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q) = A0 × R0 = 1 Thus, the given result is proved.

Q23	If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.
Ans:	The first term of the G.P is a and the last term is b. Therefore, the G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio. b = ar^n–1 … (1) P = Product of n terms = (a) (ar) (ar²) … (ar^n–1) = (a × a ×…a) (r × r² × …r^n–1) = an r ^{1 + 2 +…(n–1)}… (2) Here, 1, 2, …(n – 1) is an A.P. ∴1 + 2 + ……….+ (n – 1) $equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus left parenthesis n minus 1 minus 1 right parenthesis cross times 1 close square brackets equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus n minus 2 close square brackets equals fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction P space equals space a to the power of n r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction end exponent therefore space P squared space equals space a to the power of 2 n end exponent space r to the power of n open parentheses n minus 1 close parentheses end exponent space space space space space space space space space space space space equals space open square brackets a squared space r to the power of left parenthesis n minus 1 right parenthesis end exponent close square brackets to the power of n space space space space space space space space space space space space equals open square brackets a cross times a r to the power of n minus 1 end exponent close square brackets to the power of n space space space space space space space space space space space space equals open parentheses a b close parentheses to the power of n space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets$ Thus, the given result is proved.

Q24	Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $open parentheses n plus 1 close parentheses to the power of t h end exponent space t o space open parentheses 2 n close parentheses to the power of t h end exponent space t e r m space i s space 1 over r to the power of n$
Ans:	Let a be the first term and r be the common ratio of the G.P. $S u m space o f space f i r s t space n space t e r m s space equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction$ Since there are n terms from (n +1)th to (2n)th term, Sum of terms from(n + 1)th to (2n)th term = $fraction numerator a subscript n plus 1 end subscript open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction$ a _n +1 = ar ^{n + 1 – 1} = arⁿ Thus, required ratio = $fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction cross times fraction numerator open parentheses 1 minus r close parentheses over denominator a r to the power of n open parentheses 1 minus r to the power of n close parentheses end fraction equals 1 over r to the power of n$ Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is $1 over r to the power of n$ .

Q25	If a, b, c and d are in G.P. show that $open parentheses a squared plus b squared plus c squared close parentheses open parentheses b squared plus c squared plus d squared close parentheses equals open parentheses a b plus b c plus c d close parentheses squared.$
Ans:	a, b, c, d are in G.P. Therefore, bc = ad … (1) b2 = ac … (2) c2 = bd … (3) It has to be proved that, (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2 R.H.S. = (ab + bc + cd)2 = (ab + ad + cd)2 [Using (1)] = [ab + d (a + c)]2 = a2b2 + 2abd (a + c) + d2 (a + c)2 = a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2) = a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)] = a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2 = a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2 [Using (2) and (3) and rearranging terms] = a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2) = (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S. ∴ L.H.S. = R.H.S. ∴ $open parentheses a squared plus b squared plus c squared close parentheses open parentheses b squared plus c squared plus d squared close parentheses equals open parentheses a b plus b c plus c d close parentheses squared$

Q26	Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Ans:	Let G₁ and G₂ be two numbers between 3 and 81 such that the series, 3, G₁, G₂, 81, forms a G.P. Let a be the first term and r be the common ratio of the G.P. ∴81 = (3) (r)³ ⇒ r³ = 27 ∴ r = 3 (Taking real roots only) For r = 3, G₁ = ar = (3) (3) = 9 G₂ = ar² = (3) (3)² = 27 Thus, the required two numbers are 9 and 27.

Q27	Find the value of n so that $fraction numerator a to the power of n plus 1 end exponent space plus space b to the power of n plus 1 end exponent over denominator a to the power of n space plus space b to the power of n end fraction$ may be the geometric mean between a and b.
Ans:	G. M. of a and b is $square root of a b end root$ . By the given condition, $fraction numerator a to the power of n plus 1 end exponent space plus space b to the power of n plus 1 end exponent over denominator a to the power of n space plus space b to the power of n end fraction equals square root of a b end root$ Squaring both sides, we obtain $open parentheses a to the power of n plus 1 end exponent space plus space b to the power of n plus 1 end exponent close parentheses squared over open parentheses a to the power of n space plus space b to the power of n close parentheses squared equals a b rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space open parentheses a b close parentheses open parentheses a to the power of 2 n end exponent space plus space 2 a to the power of n b to the power of n space plus space b to the power of 2 n end exponent close parentheses rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space a to the power of 2 n plus 1 end exponent b space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space a b to the power of 2 n plus 1 end exponent rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space a to the power of 2 n plus 1 end exponent b space plus space a b to the power of 2 n plus 1 end exponent rightwards double arrow a to the power of 2 n plus 2 end exponent space minus a to the power of 2 n plus 1 end exponent b space equals space a b to the power of 2 n plus 1 end exponent space minus space b to the power of 2 n plus 2 end exponent rightwards double arrow space a to the power of 2 n plus 1 end exponent open parentheses a minus b close parentheses equals b to the power of 2 n plus 1 end exponent open parentheses a minus b close parentheses rightwards double arrow open parentheses a over b close parentheses to the power of 2 n plus 1 end exponent space equals space 1 space equals space open parentheses a over b close parentheses to the power of 0 rightwards double arrow 2 n space plus 1 space equals space 0 rightwards double arrow n space equals space fraction numerator minus 1 over denominator 2 end fraction$

Q28	The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $open parentheses 3 space plus space 2 square root of 2 close parentheses space : space open parentheses 3 minus 2 square root of 2 close parentheses.$
Ans:	Let the two numbers be a and b. G.M. = $square root of a b end root$ According to the given condition, $a space plus space b space equals space 6 square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis rightwards double arrow open parentheses a space plus space b close parentheses squared space equals space 36 open parentheses a b close parentheses$ Also, $open parentheses a minus b close parentheses squared space equals space open parentheses a plus b close parentheses squared space minus space 4 a b space equals space 36 a b space minus 4 a b space equals space 32 a b rightwards double arrow a minus b space equals space square root of 32 space square root of a b end root space equals space 4 space square root of 2 space square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$ Adding (1) and (2), we obtain $2 a space equals space open parentheses 6 space plus space 4 square root of 2 close parentheses square root of a b end root rightwards double arrow a space equals space open parentheses 3 space plus space 2 square root of 2 close parentheses square root of a b end root$ Substituting the value of a in (1), we obtain $b space equals space 6 square root of a b end root space minus space open parentheses 3 plus 2 square root of 2 close parentheses square root of a b end root rightwards double arrow b space equals space open parentheses 3 space minus space 2 square root of 2 close parentheses space square root of a b end root a over b space equals space fraction numerator open parentheses 3 space plus 2 square root of 2 close parentheses square root of a b end root over denominator open parentheses 3 minus 2 square root of 2 close parentheses square root of a b end root end fraction space equals fraction numerator 3 space plus space 2 square root of 2 over denominator 3 minus 2 square root of 2 end fraction space$ Thus, the required ratio is $open parentheses 3 space plus space 2 square root of 2 close parentheses space : space open parentheses 3 minus 2 square root of 2 close parentheses$ .

Q29	If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A plus-or-minus square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root$ .
Ans:	It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b. ∴ $A M space equals space A space equals space fraction numerator a space plus space b over denominator 2 end fraction space space space space space space space space space space space space... left parenthesis 1 right parenthesis G M space equals space G space equals space square root of a b end root space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$ From (1) and (2), we obtain a + b = 2A … (3) ab = G₂ … (4) Substituting the value of a and b from (3) and (4) in the identity (a – b)² = (a + b)² – 4ab, we obtain (a – b)² = 4A₂ – 4G₂ = 4 (A₂–G₂) (a – b)² = 4 (A + G) (A – G) $open parentheses a minus b close parentheses space equals space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root space space space space space space space space space space... left parenthesis 5 right parenthesis$ From (3) and (5), we obtain $2 a space equals space 2 A space plus space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root rightwards double arrow a space equals space A space plus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root$ Substituting the value of a in (3), we obtain $b space equals space 2 A space minus A space minus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root space equals space A space minus space square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root$ Thus, the two numbers are $A plus-or-minus square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root$ .

Q31	What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Ans:	The amount deposited in the bank is Rs 500. At the end of first year, amount = $R s. space 500 open parentheses 1 space plus space 1 over 10 close parentheses$ = Rs 500 (1.1) At the end of 2nd year, amount = Rs 500 (1.1) (1.1) At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on ∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times) = Rs 500(1.1)¹⁰

Q32	If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Ans:	Let the root of the quadratic equation be a and b. According to the given condition, $A. M. space equals fraction numerator a plus b over denominator 2 end fraction equals 8 rightwards double arrow a plus b space equals space 16 space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis G. M. space equals space square root of a b end root space equals space 5 space rightwards double arrow a b space equals space 25 space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$ The quadratic equation is given by, x²– x (Sum of roots) + (Product of roots) = 0 x² – x (a + b) + (ab) = 0 x² – 16x + 25 = 0 [Using (1) and (2)] Thus, the required quadratic equation is x² – 16x + 25 = 0

Exercise 4

Q1	Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Ans:	The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … nth term, a_n = n ( n + 1) ∴ $S subscript n space equals space sum from k equals 1 to n of a subscript k space equals space sum from k equals 1 to n of k open parentheses k plus 1 close parentheses space space space space space equals space sum from k equals 1 to n of space k squared space plus space sum from k equals 1 to n of space k space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses fraction numerator 2 n plus 1 over denominator 3 end fraction space plus space 1 close parentheses space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses fraction numerator 2 n plus 4 over denominator 3 end fraction close parentheses space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses over denominator 3 end fraction$

Q2	Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Ans:	The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … nth term, an = n ( n + 1) ( n + 2) = (n² + n) (n + 2) = n³ + 3n² + 2n ∴ $S subscript n space equals space sum from k equals 1 to n of space a subscript k space space space space space space space space space space space space equals space sum from k equals 1 to n of space k cubed space plus space 3 space sum from k equals 1 to n of space k squared space plus space 2 sum from k equals 1 to n of space k space space space space space space equals space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets squared space plus space fraction numerator 3 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus space fraction numerator 2 n open parentheses n plus 1 close parentheses over denominator 2 end fraction space space space space space space equals space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets squared space space plus space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 2 end fraction space plus space n open parentheses n plus 1 close parentheses space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus space 2 n space plus 1 plus 2 close square brackets space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open square brackets fraction numerator n squared space plus n plus 4 n plus 6 over denominator 2 end fraction close square brackets space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 4 end fraction open parentheses n squared space plus 5 n space plus 6 close parentheses space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 4 end fraction open parentheses n squared space plus space 2 n space plus space 3 n space plus 6 close parentheses space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open square brackets n open parentheses n plus 2 close parentheses plus 3 open parentheses n plus 2 close parentheses close square brackets over denominator 4 end fraction space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses open parentheses n plus 3 close parentheses over denominator 4 end fraction$

Q3	Find the sum to n terms of the series 3 × 1² + 5 × 2² + 7 × 3² + …
Ans:	The given series is 3 ×1² + 5 × 2² + 7 × 3² + … nth term, a_n = ( 2n + 1) n² = 2n³ + n² ^{$therefore space S subscript n space end subscript equals space sum from k equals 1 to n of space a subscript k space space space space space space space space space space space space equals space sum from k equals 1 to n of space open parentheses 2 k cubed space plus space k squared close parentheses space equals space 2 sum from k equals 1 to n of space k cubed space plus space sum from k equals 1 to n of space k squared space space space space space space space space space space space space equals space 2 open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets squared space plus space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space space space space space space space space space space space space equals space fraction numerator n squared open parentheses n plus 1 close parentheses squared over denominator 2 end fraction space plus space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space space space space space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open square brackets n open parentheses n plus 1 close parentheses space plus space fraction numerator 2 n plus 1 over denominator 3 end fraction close square brackets space space space space space space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space open square brackets fraction numerator 3 n squared space plus space 3 n space plus space 2 n plus space 1 over denominator 3 end fraction close square brackets space space space space space space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space open square brackets fraction numerator 3 n squared space plus space 5 n space plus 1 over denominator space 3 end fraction close square brackets space space space space space space space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses 3 n squared space plus space 5 n space plus 1 close parentheses over denominator 6 end fraction$}

Q4	Find the sum to n terms of the series $fraction numerator 1 over denominator 1 cross times 2 end fraction space plus fraction numerator 1 over denominator 2 cross times 3 end fraction space plus space fraction numerator 1 over denominator 3 space cross times space 4 end fraction space plus space...$
Ans:	The given series is $fraction numerator 1 over denominator 1 cross times 2 end fraction space plus space fraction numerator 1 over denominator 2 cross times 3 end fraction space plus fraction numerator 1 over denominator 3 cross times 4 end fraction space plus space...$ nth term, a_n = $fraction numerator 1 over denominator n open parentheses n plus 1 close parentheses end fraction equals 1 over n minus fraction numerator 1 over denominator n plus 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis B y space p a r t i a l space f r a c t i o n s right parenthesis$ $a subscript 1 space equals space 1 over 1 space minus space 1 half a subscript 2 space equals space 1 half space minus space 1 third a subscript 3 space space equals space 1 third space minus space 1 fourth... a subscript n space equals space 1 over n space minus space fraction numerator 1 over denominator n plus 1 end fraction$ Adding the above terms column wise, we obtain $a subscript 1 space plus space a subscript 2 space plus space... space plus space a subscript n space equals space open square brackets 1 over 1 plus 1 half plus 1 third plus...1 over n close square brackets minus open square brackets 1 half plus 1 third plus 1 fourth plus... fraction numerator 1 over denominator n plus 1 end fraction close square brackets therefore space S subscript n space equals space 1 minus space fraction numerator 1 over denominator n plus 1 end fraction equals fraction numerator n plus 1 minus 1 over denominator n plus 1 end fraction equals fraction numerator n over denominator n plus 1 end fraction$

Q5	Find the sum to n terms of the series 5² + 6² + 7² + ... + 20²
Ans:	The given series is 5² + 6² + 7² + … + 20² nth term, a_n = ( n + 4)² = n² + 8n + 16 $therefore space S subscript n space equals space sum from k equals 1 to n of space a subscript k space equals space sum from k equals 1 to n of space open parentheses k squared space plus space 8 k space plus space 16 close parentheses space space space space space space space space space space space space equals space sum from k equals 1 to n of space k squared space plus space 8 sum from k equals 1 to n of space k space plus space sum from k equals 1 to n of space 16 space space space space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus fraction numerator 8 n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus 16 n$ 16th term is (16 + 4)² = (20)² ^{$therefore space S subscript n space equals space fraction numerator 16 open parentheses 16 plus 1 close parentheses open parentheses 2 cross times 16 plus 1 close parentheses over denominator 6 end fraction plus fraction numerator 8 cross times 16 cross times open parentheses 16 plus 1 close parentheses over denominator 2 end fraction plus 16 cross times 16 space space space space space space space space space space space equals space space fraction numerator open parentheses 16 close parentheses open parentheses 17 close parentheses open parentheses 33 close parentheses over denominator 6 end fraction plus fraction numerator open parentheses 8 close parentheses cross times 16 cross times open parentheses 16 plus 1 close parentheses over denominator 2 end fraction plus 16 cross times 16 space space space space space space space space space space space equals space space fraction numerator open parentheses 16 close parentheses open parentheses 17 close parentheses open parentheses 33 close parentheses over denominator 6 end fraction plus fraction numerator open parentheses 8 close parentheses open parentheses 16 close parentheses open parentheses 17 close parentheses over denominator 2 end fraction plus 256 space space space space space space space space space space equals 1496 space plus space 1088 space plus space 256 space space space space space space space space space space equals space 2840 therefore space 5 squared space plus space 6 to the power of 2 space end exponent plus space 7 squared space plus space...... space plus space 20 squared space equals space 2840$}

Q6	Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…
Ans:	The given series is 3 × 8 + 6 × 11 + 9 × 14 + … an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …) = (3n) (3n + 5) = 9n² + 15n $therefore space S subscript n space equals space sum from k equals 1 to n of space a subscript k space equals space sum from k equals 1 to n of space open parentheses 9 k squared space plus space 15 k close parentheses space space space space space space space space space space space equals space 9 sum from k equals 1 to n of space k squared space plus space 15 space sum from k equals 1 to n of space k space space space space space space space space space space space equals space 9 space cross times space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction plus 15 cross times fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space space space space space space space space space space space space equals fraction numerator 3 n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses 2 n plus 1 plus 5 close parentheses space space space space space space space space space space space space equals fraction numerator 3 n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses 2 n plus 6 close parentheses space space space space space space space space space space space space space space equals space 3 n open parentheses n plus 1 close parentheses open parentheses n plus 3 close parentheses$

Q7	Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …
Ans:	The given series is 1² + (1² + 2²) + (1² + 2² + 3³ ) + … a_n = (1² + 2² + 3³ +…….+ n²) $equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction equals fraction numerator n open parentheses 2 n squared space plus space 3 n space plus 1 close parentheses over denominator 6 end fraction space equals space fraction numerator 2 cubed space plus space 3 n squared space plus space n over denominator 6 end fraction equals space 1 third n cubed space plus space 1 half space n squared space plus space 1 over 6 space n therefore space S subscript n space equals space sum from k equals 1 to n of space a subscript k space space space space space space space space space space equals sum from k equals 1 to n of space open parentheses 1 third space k cubed space plus space 1 half space k squared space plus space 1 over 6 space k close parentheses space space space space space space space space space space equals space 1 third space sum from k equals 1 to n of space k cubed space plus space 1 half space sum from k equals 1 to n of space k squared space plus space 1 over 6 space sum from k equals 1 to n of space k space space space space space space space space space equals 1 third fraction numerator n squared space open parentheses n plus 1 close parentheses squared over denominator open parentheses 2 close parentheses squared end fraction space plus space 1 half space cross times space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus space 1 over 6 space cross times fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 6 end fraction space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus space fraction numerator open parentheses 2 n plus 1 close parentheses over denominator 2 end fraction plus 1 half close square brackets space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 6 end fraction space open square brackets fraction numerator n squared space plus space n space plus space 2 n space plus 1 space plus 1 over denominator 2 end fraction close square brackets space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 6 end fraction space open square brackets fraction numerator n open parentheses n plus 1 close parentheses plus 2 open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 6 end fraction space open square brackets fraction numerator open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses over denominator 2 end fraction close square brackets space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses squared open parentheses n plus 2 close parentheses over denominator 12 end fraction$

Q8	Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Ans:	a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n $therefore space S subscript n space equals space sum from k equals 1 to n of space a subscript k space space end subscript space space space space space space space space space space space space equals sum from k equals 1 to n of space open parentheses k cubed space plus space 5 k squared space plus space 4 k close parentheses space space space space space space space space space space space equals space sum from k equals 1 to n of space k cubed space plus space 5 space sum from k equals 1 to n of space k squared space plus space 4 sum from k equals 1 to n of space k space space space space space space space space space space space equals space fraction numerator n squared open parentheses n plus 1 close parentheses squared over denominator open parentheses 2 close parentheses squared end fraction space plus space fraction numerator 5 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus space fraction numerator 4 n open parentheses n plus 1 close parentheses over denominator 2 end fraction space space space space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus space fraction numerator 5 open parentheses 2 n plus 1 close parentheses over denominator 3 end fraction space plus space 4 close square brackets space space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space open square brackets fraction numerator 3 n squared space plus space 3 n space plus space 20 n space plus space 10 space plus 24 over denominator 6 end fraction close square brackets space space space space space space space space space space equals space space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space open square brackets fraction numerator 3 n squared space plus space 23 n space plus space 34 over denominator 6 end fraction close square brackets space space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses 3 n squared space plus space 23 n space plus 34 close parentheses over denominator 12 end fraction$

Q9	Find the sum to n terms of the series whose nth terms is given by n² + 2n
Ans:	a_n = n² + 2n $therefore space S subscript n space equals space sum from k equals 1 to n of space k squared space plus space 2 to the power of k space equals space sum from k equals 1 to n of space k squared space space plus space sum from k equals 1 to n of space 2 to the power of k space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis C o n s i d e r space sum from k equals 1 to n of space 2 to the power of k space equals space 2 to the power of 1 space plus space 2 squared space plus 2 cubed space plus space...$ The above series 2, 2², 2³, … is a G.P. with both the first term and common ratio equal to 2. $therefore space sum from k equals 1 to n of space 2 to the power of k space equals space fraction numerator open parentheses 2 close parentheses open square brackets open parentheses 2 close parentheses to the power of n space minus 1 close square brackets over denominator 2 minus 1 end fraction equals 2 open parentheses 2 to the power of n space minus 1 close parentheses space space space space space space space space space space space... left parenthesis 2 right parenthesis$ Therefore, from (1) and (2), we obtain $S subscript n space equals space sum from k equals 1 to n of space k squared space plus space 2 open parentheses 2 to the power of n space minus 1 close parentheses space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction plus 2 open parentheses 2 to the power of n space end exponent minus 1 close parentheses$

Q10	Find the sum to n terms of the series whose nth terms is given by (2n – 1)²
Ans:	a_n = (2n – 1)² = 4n² – 4n + 1 $therefore space S subscript n space end subscript equals space sum from k equals 1 to n of space a subscript k space equals space sum from k equals 1 to n of space open parentheses 4 k squared space minus 4 k space plus 1 close parentheses space space space space space space space space space space space equals space 4 sum from k equals 1 to n of space k squared space minus space 4 sum from k equals 1 to n of space k space plus space sum from k equals 1 to n of space 1 space space space space space space space space space space space equals space fraction numerator 4 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space minus space fraction numerator 4 n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus space n space space space space space space space space space space space equals space fraction numerator 2 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 3 end fraction space minus space 2 n open parentheses n plus 1 close parentheses space plus n space space space space space space space space space space space space equals space n open square brackets fraction numerator 2 open parentheses 2 n squared space plus 3 n space plus 1 close parentheses over denominator 3 end fraction space minus space 2 n open parentheses n plus 1 close parentheses space plus 1 close square brackets space space space space space space space space space space space equals space n open square brackets fraction numerator 4 n squared space plus space 6 n space plus space 2 space minus 6 n space minus 6 space plus 3 over denominator 3 end fraction close square brackets space space space space space space space space space space space space equals space n open square brackets fraction numerator 4 n squared space minus 1 over denominator 3 end fraction close square brackets space space space space space space space space space space space equals space fraction numerator n open parentheses 2 n plus 1 close parentheses open parentheses 2 n minus 1 close parentheses over denominator 3 end fraction$

Exercise 5

Q1	Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Ans:	Let a and d be the first term and the common difference of the A.P. respectively. It is known that the kth term of an A. P. is given by ak = a + (k –1) d ∴ am + n = a + (m + n –1) d am – n = a + (m – n –1) d am = a + (m –1) d ∴ am + n + am – n = a + (m + n –1) d + a + (m – n –1) d = 2a + (m + n –1 + m – n –1) d = 2a + (2m – 2) d = 2a + 2 (m – 1) d =2 [a + (m – 1) d] = 2am Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Q2	If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Ans:	Let the three numbers in A.P. be a – d, a, and a + d. According to the given information, (a – d) + (a) + (a + d) = 24 … (1) ⇒ 3a = 24 ∴ a = 8 (a – d) a (a + d) = 440 … (2) ⇒ (8 – d) (8) (8 + d) = 440 ⇒ (8 – d) (8 + d) = 55 ⇒ 64 – d2 = 55 ⇒ d2 = 64 – 55 = 9 ⇒ d = ± 3 Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5. Thus, the three numbers are 5, 8, and 11.

Q3	Let the sum of n, 2n, 3n terms of an A.P. be S₁, S₂ and S₃, respectively, show that S₃ = 3 (S₂– S₁)
Ans:	Let a and b be the first term and the common difference of the A.P. respectively. Therefore, $S subscript 1 space equals space n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis S subscript 2 space equals space fraction numerator 2 n over denominator 2 end fraction open square brackets 2 a plus open parentheses 2 n minus 1 close parentheses d close square brackets space equals space n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space space... left parenthesis 2 right parenthesis S subscript 3 space equals space fraction numerator 3 n over denominator 2 end fraction open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets space space space space space... left parenthesis 3 right parenthesis$ From (1) and (2), we obtain $S subscript 2 space minus space S subscript 1 space equals n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space minus n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets space space space space space space space space space space space space space space space space equals space n open curly brackets fraction numerator 4 a space plus space 4 n d space minus 2 d space minus 2 a space minus n d space plus d over denominator 2 end fraction close curly brackets space space space space space space space space space space space space space space space space equals space n open square brackets fraction numerator 2 a space plus space 3 n d space minus d over denominator 2 end fraction close square brackets space space space space space space space space space space space space space space space equals space n over 2 space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets therefore space 3 open parentheses S subscript 2 space minus space S subscript 1 close parentheses space equals space fraction numerator 3 n over denominator 2 end fraction space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets equals S subscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets F r o m space left parenthesis 3 right parenthesis close square brackets$ Hence, the given result is proved.

Q4	Find the sum of all numbers between 200 and 400 which are divisible by 7.
Ans:	The numbers lying between 200 and 400, which are divisible by 7, are 203, 210, 217, … 399 ∴First term, a = 203 Last term, l = 399 Common difference, d = 7 Let the number of terms of the A.P. be n. ∴ a_n = 399 = a + (n –1) d ⇒ 399 = 203 + (n –1) 7 ⇒ 7 (n –1) = 196 ⇒ n –1 = 28 ⇒ n = 29 $therefore space S subscript 29 space equals space 29 over 2 space open parentheses 203 space plus space 399 close parentheses space space space space space space space space space space space space space equals space 29 over 2 space open parentheses 602 close parentheses space space space space space space space space space space space space space equals space open parentheses 29 close parentheses open parentheses 301 close parentheses space space space space space space space space space space space space space equals space 8729$ Thus, the required sum is 8729.

Q5	Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Ans:	The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100. This forms an A.P. with both the first term and common difference equal to 2. ⇒100 = 2 + (n –1) 2 ⇒ n = 50 $therefore space 2 plus 4 plus 6 plus... plus 100 space equals 50 over 2 space open square brackets 2 open parentheses 2 close parentheses space plus space open parentheses 50 minus 1 close parentheses open parentheses 2 close parentheses close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 over 2 space open square brackets 4 space plus space 98 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 25 close parentheses open parentheses 102 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2550$ The integers from 1 to 100, which are divisible by 5, are 5, 10… 100. This forms an A.P. with both the first term and common difference equal to 5. ∴100 = 5 + (n –1) 5 ⇒ 5n = 100 ⇒ n = 20 $therefore space 5 space plus space 10 space plus space... space plus space 100 space equals space 20 over 2 space open square brackets 2 open parentheses 5 close parentheses space plus space open parentheses 20 space minus 1 close parentheses 5 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space open parentheses 19 close parentheses 5 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space 95 close square brackets space equals space 10 space cross times space 105 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1050$ The integers, which are divisible by both 2 and 5, are 10, 20, … 100. This also forms an A.P. with both the first term and common difference equal to 10. ∴100 = 10 + (n –1) (10) ⇒ 100 = 10n ⇒ n = 10 $therefore space 10 space plus space 20 space plus space... space plus space 100 space equals space 10 over 2 space open square brackets 2 open parentheses 10 close parentheses space plus space open parentheses 10 space minus 1 close parentheses open parentheses 10 close parentheses close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space open square brackets 20 space plus space 90 close square brackets space equals space 5 open parentheses 110 close parentheses space equals space 550$ ∴Required sum = 2550 + 1050 – 550 = 3050 Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Q6	Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Ans:	The two-digit numbers, which when divided by 4, yield 1 as remainder, are 13, 17, … 97. This series forms an A.P. with first term 13 and common difference 4. Let n be the number of terms of the A.P. It is known that the nth term of an A.P. is given by, an = a + (n –1) d ∴97 = 13 + (n –1) (4) ⇒ 4 (n –1) = 84 ⇒ n – 1 = 21 ⇒ n = 22 Sum of n terms of an A.P. is given by, $S subscript n space equals space n over 2 space open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets therefore space S subscript 22 space equals space 22 over 2 space open square brackets 22 open parentheses 13 close parentheses space plus space open parentheses 22 minus 1 close parentheses open parentheses 4 close parentheses close square brackets space space space space space space space space space space space space space equals space 11 open square brackets 26 space plus space 84 close square brackets space space space space space space space space space space space space space equals space 1210$ Thus, the required sum is 1210.

Q7	If f is a function satisfying f(x +y) = f(x) f(y) for all x,y $element of$ N such that f(1) = 3 $sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120$ and , find the value of n.
Ans:	It is given that, f (x + y) = f (x) × f (y) for all x, y ∈ N … (1) f (1) = 3 Taking x = y = 1 in (1), we obtain f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9 Similarly, f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27 f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81 ∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3. It is known that, $S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r space minus 1 end fraction$ It is given that, $sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120$ $therefore space 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction rightwards double arrow 120 space equals space 3 over 2 open parentheses 3 to the power of n space minus space 1 close parentheses rightwards double arrow 3 to the power of n space minus space 1 space equals space 80 rightwards double arrow 3 to the power of n space equals space 81 space equals space 3 to the power of 4 space therefore space n space equals 4$ Thus, the value of n is 4.

Q8	The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Ans:	Let the sum of n terms of the G.P. be 315. It is known that, $S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction$ It is given that the first term a is 5 and common ratio r is 2. $therefore space 315 space equals fraction numerator 5 open parentheses 2 to the power of n minus 1 close parentheses space over denominator 2 minus 1 end fraction rightwards double arrow 2 to the power of n space end exponent minus 1 space equals space 63 rightwards double arrow 2 to the power of n space equals 64 space equals space open parentheses 2 close parentheses to the power of 6 rightwards double arrow n equals 6$ ∴Last term of the G.P = 6th term = ar⁶ ^{– 1} = (5)(2)⁵ = (5)(32) = 160 Thus, the last term of the G.P. is 160.

Q9	The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Ans:	Let a and r be the first term and the common ratio of the G.P. respectively. ∴ a = 1 a₃ = ar² = r2 a₅ = ar⁴ = r4 ∴ r² + r⁴ = 90 ⇒ r⁴ + r² – 90 = 0 $rightwards double arrow r squared space equals space fraction numerator minus 1 space plus space square root of 1 plus space 360 end root over denominator 2 end fraction equals fraction numerator minus 1 space plus-or-minus square root of 361 over denominator 2 end fraction equals fraction numerator minus 1 plus-or-minus 19 over denominator 2 end fraction equals minus 10 space o r space 9$ ∴ r = ± 3 ( Taking real roots) Thus, the common ratio of the G.P. is ±3.

Q10	The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Ans:	Let the three numbers in G.P. be a, ar, and ar2. From the given condition, a + ar + ar2 = 56 ⇒ a (1 + r + r2) = 56 $rightwards double arrow a space equals fraction numerator 56 over denominator 1 plus r plus r squared end fraction space space space space... space left parenthesis 1 right parenthesis$ a – 1, ar – 7, ar2 – 21 forms an A.P. ∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7) ⇒ ar – a – 6 = ar2 – ar – 14 ⇒ar2 – 2ar + a = 8 ⇒ar2 – ar – ar + a = 8 ⇒a(r2 + 1 – 2r) = 8 ⇒ a (r – 1)2 = 8 … (2) $rightwards double arrow fraction numerator 56 over denominator 1 plus r plus r squared end fraction open parentheses r minus 1 close parentheses squared equals 8 space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets$ ⇒7(r2 – 2r + 1) = 1 + r + r2 ⇒7r2 – 14 r + 7 – 1 – r – r2 = 0 ⇒ 6r2 – 15r + 6 = 0 ⇒ 6r2 – 12r – 3r + 6 = 0 ⇒ 6r (r – 2) – 3 (r – 2) = 0 ⇒ (6r – 3) (r – 2) = 0 $therefore space r space equals space 2 comma 1 half$ When r = 2, a = 8 When $r space equals space 1 half space comma space a space equals space 32$ Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32. When $r space equals space 1 half$ , the three numbers in G.P. are 32, 16, and 8. Thus, in either case, the three required numbers are 8, 16, and 32.

Q11	A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Ans:	Let the G.P. be T₁, T₂, T₃, T₄, … T_2n. Number of terms = 2n According to the given condition, T₁ + T₂ + T₃ + …+ T_2n = 5 [T₁ + T₃ + … +T_2n–1] ⇒ T₁ + T₂ + T₃ + … + T_{2n – 5} [T₁ + T₃ + … + T_2n–1] = 0 ⇒ T₂ + T₄ + … + T_2n = 4 [T₁ + T₃ + … + T_2n–1] Let the G.P. be a, ar, ar², ar³, … $therefore space fraction numerator a r space open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction space equals space fraction numerator 4 space cross times space a open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction rightwards double arrow a r space equals space 4 a rightwards double arrow r space equals space 4$ Thus, the common ratio of the G.P. is 4.

Q12	The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Ans:	Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d. Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] = 4a + (4n – 10) d According to the given condition, 4a + 6d = 56 ⇒ 4(11) + 6d = 56 [Since a = 11 (given)] ⇒ 6d = 12 ⇒ d = 2 ∴ 4a + (4n –10) d = 112 ⇒ 4(11) + (4n – 10)2 = 112 ⇒ (4n – 10)2 = 68 ⇒ 4n – 10 = 34 ⇒ 4n = 44 ⇒ n = 11 Thus, the number of terms of the A.P. is 11.

Q13	If $fraction numerator a space plus b x over denominator a space minus space b x end fraction space equals space fraction numerator b space plus space c x over denominator b space minus space c x end fraction space equals space fraction numerator c space plus space d x over denominator c space minus d x end fraction space open parentheses x space not equal to 0 close parentheses$ Ans:	It is given that, $fraction numerator a space plus space b x over denominator a space minus b x end fraction space equals space fraction numerator b space plus space c x over denominator b space minus c x end fraction rightwards double arrow space open parentheses a space plus space b x close parentheses open parentheses b minus c x close parentheses space equals space open parentheses b plus c x close parentheses open parentheses a minus b x close parentheses rightwards double arrow a b space minus space a c x space plus space b squared x space minus space b c x squared space equals space a b space minus space b squared x space plus space a c x space minus space b c x squared rightwards double arrow 2 b squared x space equals space 2 a c x rightwards double arrow b squared space equals space a c rightwards double arrow b over a space equals space c over b space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis A l s o comma space fraction numerator b plus c x over denominator b minus c x end fraction space equals space fraction numerator c space plus d x over denominator c space minus d x end fraction space rightwards double arrow open parentheses b plus c x close parentheses open parentheses c minus d x close parentheses space equals space open parentheses b minus c x close parentheses open parentheses c plus d x close parentheses rightwards double arrow b c space minus space b d x space plus space c squared x space minus space c d x squared space equals space b c space plus space b d x space minus space c squared x space minus space c d x squared rightwards double arrow 2 c squared x space equals space 2 b d x rightwards double arrow c squared space equals space b d rightwards double arrow c over d space equals space d over c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$ From (1) and (2), we obtain $b over a space equals space c over b space equals d over c$ Thus, a, b, c, and d are in G.P.

Q14	Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ
Ans:	Let the G.P. be a, ar, ar², ar³, … ar^{n – 1}… According to the given information, $S space equals space fraction numerator a space open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction P space equals space a to the power of n space end exponent cross times space r to the power of 1 plus 2 plus... plus n minus 1 end exponent space space space space equals space a to the power of n space end exponent r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction space space space space space space space open square brackets because space S u m space o f space f i r s t space n space n a t u r a l space n u m b e r s space i s space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets end exponent R space equals space 1 over a space plus space fraction numerator 1 over denominator a r end fraction space plus space... space plus space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction space space space space equals space fraction numerator r to the power of n minus 1 end exponent plus r to the power of n minus 2 end exponent space plus space... r plus 1 over denominator a r to the power of n minus 1 end exponent end fraction space space space space equals space fraction numerator 1 open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction space cross times space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction space space space space space space space space space space open square brackets because space 1 comma r comma... r to the power of n minus 1 end exponent space f o r m s space a space G. P. close square brackets space space space space equals space fraction numerator r to the power of n space minus 1 over denominator a r to the power of n minus 1 end exponent open parentheses r minus 1 close parentheses end fraction therefore space P squared R to the power of n space equals space a to the power of 2 n end exponent r to the power of n open parentheses n minus 1 close parentheses space end exponent fraction numerator open parentheses r to the power of n space minus 1 close parentheses to the power of n over denominator a to the power of n r to the power of n open parentheses n minus 1 close parentheses end exponent open parentheses r minus 1 close parentheses to the power of n end fraction space space space space space space space space space space space space space space space space space equals space fraction numerator a to the power of n open parentheses r to the power of n minus 1 close parentheses to the power of n over denominator open parentheses r minus 1 close parentheses to the power of n end fraction space space space space space space space space space space space space space space space space space equals space open square brackets fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction close square brackets to the power of n space space space space space space space space space space space space space space space space equals space S to the power of n$ Hence P² Rⁿ= S_n