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

Answer 1

Given: $\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=31$
$\Rightarrow \frac{\frac{3n}{2}[2a+(3n-1\left)d\right]-\frac{n-1}{2}[2a+(n-2\left)d\right]}{T_{2n}}=31(\because S_{2n}-S_{2n-1}=T_{2n})\Rightarrow \frac{2a(3n-(n-1\left)\right)+d\left(3n\right(3n-1)-(n-1\left)\right(n-2\left)\right)}{2T_{2n}}=31\Rightarrow \frac{2a(2n+1)+d(8n^2-2)}{2T_{2n}}=31\Rightarrow \frac{2(2n+1)[a+(2n-1\left)d\right]}{2T_{2n}}=31\Rightarrow \frac{(2n+1)T_{2n}}{T_{2n}}=31\Rightarrow 2n+1=31\therefore n=15$