Question

Let $S_n$ denote the sum of first n terms of an AP, if $S_{2n}=3S_n,$ then the ratio $\frac{S_{3n}}{S_n}$ is equal to

• A. 4
• B. 6
• C. $\pm 3$
• D. $\pm 4$

Answer 1

Answer: (B)

Given, $S_{2n}=3S_n$
$\Rightarrow$ $\frac{2n}{2}[2a+(2n-1)d]=3.\frac{n}{2}[2a+(n-1)d]$
where a is the first term and d is the common difference of given AP.
$\Rightarrow$ $2a=(n+1)d$
$\therefore$ $\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}$
$=\frac{12nd}{2nd}=6$