First slide

Arithmetic progression

Question

If S_n denotes the sum of first n terms of an A.P. and $\frac{S_{3 n} - S_{n - 1}}{S_{2 n} - S_{2 n - 1}} = 31$ , then the value of n is

Moderate

A

21
B

15
C

16
D

19

Solution

$\begin{matrix} S_{3 n} = \frac{3 n}{2} [2 a + (3 n - 1) d] \\ S_{n - 1} = \frac{n - 1}{2} [2 a + (n - 2) d] \end{matrix}$
$\begin{array}{rcl} \Rightarrow & S_{3 n} - S_{n - 1} = \frac{1}{2} [2 a (3 n - n + 1)] + \frac{d}{2} [3 n (3 n - 1) - (n - 1) (n - 2)] \\ \begin{matrix} = \frac{1}{2} [2 a (2 n + 1) + d (8 n^{2} - 2)] \\ = a (2 n + 1) + d (4 n^{2} - 1) \\ = (2 n + 1) [a + (2 n - 1) d] \end{matrix} \end{array}$
$S_{2 n} - S_{2 n - 1} = T_{2 n} = a + (2 n - 1) d$
$\Rightarrow \frac{S_{3 n} - S_{n - 1}}{S_{2 n} - S_{2 n - 1}} = (2 n + 1)$
Given,
$\begin{array}{l} \frac{S_{3 n} - S_{n - 1}}{S_{2 n} - S_{2 n - 1}} = 31 \\ \Rightarrow n = 15 \end{array}$

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