Question

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

• A. 21
• B. 15
• C. 16
• D. 19

Answer 1

The correct option is B

15

$S_{3n}=\frac{3n}{2}[2a+(3n-1\left)d\right]$

$S_{n-1}=\frac{n-1}{2}[2a+(2a+(n-2)d\left)\right]$

$\Rightarrow S_{3n}-S_{n-1}=\frac{1}{2}[2a+(3n-1\left)d\right]+\frac{d}{2}\left[3n\right(3n-1)-(n-1\left)\right(n-2)$

$=\frac{1}{2}\left[2a\right(2n+1)+d(8n^2-2)$

$=a(2n+1)+d(4n^2-2)$

$=(2n+1)[a+(2n-1\left)d\right]$

$S_{2n}-S_{2n-1}=T_{2n}=a+(2n-1)d$

$\Rightarrow \frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=(2n+1)$

Given,

$\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=31\Rightarrow n=15$