Question

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

• A. $21$
• B. $15$
• C. $16$
• D. $19$

Answer 1

Answer: (B)

$S_{3n} =\frac{3n}{2}[2a+(3n-1)d]$
$S_{n-1} =\frac{n-1}{2} [2a+(n-2)d]$
$\Rightarrow S_{3n} -S_{n-1} =\frac{1}{2} [2a(3n-n+1)]$
$+\frac{d}{2}[3n(3n-1)-(n-1)(n-2)]$
$=\frac{1}{2}[2a(2n+1)+d(8n^{2} -2)]$
$=a(2n+1) +d(4n^{2}-1)]$
$=(2n+1)[a+(2n-1)d]$
$s_{2n} -S_{2n-1} =T_{2n} =a+(2n-1)d$
$\Rightarrow \frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=(2n+1)$
Given, $\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=31 $
$\Rightarrow n=15$