Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 2, Question 9: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
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.
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Read MoreWelcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 9: The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio o....
Comments
We take 35 because. . If the ratio of the sum of n terms of ap is given, then to find the ratio of their nth terms, we replace n by (2n-1) in the ratio of the sums of n terms. 2*18-1 = 35
TO OBTAIN THE 18TH TERMS (n-1)/2=18-1 (n-1)/2=17 n-1=34 n=34+1 n=35
Why do we take n=35 in (1)?pls tell me
(n-1)/2=An-1
Why do we take n=35 in (1)?