Question

If $S_n$ denotes the sum of the 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 $15$

$S_{3n}=\frac{3n}{2}[2a+(3n-1\left)d\right]$ where $a\to$ first term $d\to$ common differ

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

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

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

$\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=\frac{\frac{3n}{2}[2a+(3n-1\left)d\right]-\frac{(n-1)}{2}[2a+(n-2\left)d\right]}{\frac{2n}{2}[2a+(2n-1\left)d\right]-\frac{(2n-1)}{2}[2a+(2n-2\left)d\right]}=31$

$=\frac{a(2n+1)+d(4n^2-1]}{a+d[2n-1)}=\frac{(2n+1)[a+(2n-1\left)d\right]}{a+(2n-1)d}=31$

$\Rightarrow (2n+1)=31$

$\Rightarrow n=15$.