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The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
nth term, an = n ( n + 1)
∴ 
an = (2n – 1)2 = 4n2 – 4n + 1
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
∴
The given series is 3 ×12 + 5 × 22 + 7 × 32 + …
nth term, an = ( 2n + 1) n2 = 2n3 + n2
The given series is 
nth term, an = 
Adding the above terms column wise, we obtain
The given series is 52 + 62 + 72 + … + 202
nth term, an = ( n + 4)2 = n2 + 8n + 16
16th term is (16 + 4)2 = (20)2
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
The given series is 12 + (12 + 22) + (12 + 22 + 33 ) + …
an = (12 + 22 + 33 +…….+ n2)
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n
an = n2 + 2n
The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.
Therefore, from (1) and (2), we obtain
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.
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
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)
⇒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
When r = 2, a = 8
When 
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When 
, the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.
Let the G.P. be T1, T2, T3, T4, … T2n.
Number of terms = 2n
According to the given condition,
T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]
⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0
⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1]
Let the G.P. be a, ar, ar2, ar3, …
Thus, the common ratio of the G.P. is 4.
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.
Let the G.P. be a, ar, ar2, ar3, … arn – 1…
According to the given information,
Hence P2 Rn= Sn
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.
Let a and b be the first term and the common difference of the A.P. respectively.
Therefore,
From (1) and (2), we obtain
Hence, the given result is proved.
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.
∴ an = 399 = a + (n –1) d
⇒ 399 = 203 + (n –1) 7
⇒ 7 (n –1) = 196
⇒ n –1 = 28
⇒ n = 29
Thus, the required sum is 8729.
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
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
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
∴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.
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,
Thus, the required sum is 1210.
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, 
It is given that, 
Thus, the value of n is 4.
Let the sum of n terms of the G.P. be 315.
It is known that, 
It is given that the first term a is 5 and common ratio r is 2.
∴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.
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
∴ r = ± 3 ( Taking real roots)
Thus, the common ratio of the G.P. is ±3.
an = 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.
\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}
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.
Let a1 and d be the first term and the common difference of the A.P. respectively.
According to the given information,
Let a and b be the first term and the common difference of the A.P. respectively.
According to the given condition,
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,
Here,
Comparing the coefficient of n2 on both sides, we obtain
Comparing the coefficient of n on both sides, we obtain
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.
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.
A.M. of a and b = 
According to the given condition,
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
A1 = a + d
A2 = a + 2d
A3 = a + 3d …
∴ A7 = a + 7d
Am–1 = a + (m – 1) d
According to the given condition,
Thus, the value of m is 14.
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.
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
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.
Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.
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.
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}
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}
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.
It is given that the kth term of the A.P. is 5k + 1.
kth term = ak = 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}
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.
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.
Here, a = First Term = 
r = common ratio = 
The given G.P. is x3, x5, x7 ........
Here, a = x3 and r = x2
Let 
be the first three terms of the G.P.
From (2), we obtain
a3 = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
The given G.P. is 3, 32, 33, …
Let n terms of this G.P. be required to obtain the sum as 120.
Here, a = 3 and r = 3
∴ 
∴ n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.
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
Substituting r = 2 in (1), we obtain
a (1 + 2 + 4) = 16
⇒ a (7) = 16
a = 729
a7 = 64
Let r be the common ratio of the G.P.
It is known that, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
Also, it is known that,
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
From (1), we obtain
.....
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
Dividing (3) by (2), we obtain
∴ 
Thus, x, y, z are in G. P.
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
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 ×
Here, 4, 2, 1, 
is a G.P.
First term, a = 4
Common ratio, r =
It is known that,
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)
It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.
Thus, the above sequence forms a G.P. and the common ratio is rR.
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
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.
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.
The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.
b = arn–1 … (1)
P = Product of n terms
= (a) (ar) (ar2) … (arn–1)
= (a × a ×…a) (r × r2 × …rn–1)
= an r 1 + 2 +…(n–1) … (2)
Here, 1, 2, …(n – 1) is an A.P.
∴1 + 2 + ……….+ (n – 1)
Thus, the given result is proved.
Let a be the first term and r be the common ratio of the G.P.
Since there are n terms from (n +1)th to (2n)th term,
Sum of terms from(n + 1)th to (2n)th term =
a n +1 = ar n + 1 – 1 = arn
Thus, required ratio = 
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 
.
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.
∴ 
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.
G. M. of a and b is 
.
By the given condition, 
Squaring both sides, we obtain
Let the two numbers be a and b.
G.M. =
According to the given condition,
Also,
Adding (1) and (2), we obtain
Substituting the value of a in (1), we obtain
Thus, the required ratio is 
.
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.
∴ 
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)
From (3) and (5), we obtain
Substituting the value of a in (3), we obtain
Thus, the two numbers are 
.
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
Dividing equation (3) by (2), we obtain
Equating the values of r3 obtained in (4) and (5), we obtain
Thus, the given result is proved.
The amount deposited in the bank is Rs 500.
At the end of first year, amount = 
= 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
Let the root of the quadratic equation be a and b.
According to the given condition,
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
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.
(a) The given sequence is 
Here, a = 2 and r = 
Let the nth term of the given sequence be 128.
Thus, the 13th term of the given sequence is 128.
(b) The given sequence is 
Here, a= 
Let the nth term of the given sequence be 729.
Thus, the 12th term of the given sequence is 729.
(c) The given sequence is 
Here,
a=
Let the nth term of the given sequence be 
.
Thus, the 9th term of the given sequence is .
The given numbers are 
.
Common ratio = 
Also, common ratio = 
so
Thus, for x = ± 1, the given numbers will be in G.P.
The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r = 
The given G.P. is 
Here, a = 
The given G.P. is 1,-a, a2,-a3, .........
Here, first term = a1 = 1
Common ratio = r = – a
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