Question

Let $S_n$ denote the sum of first n terms of an A.P. If $S_{2n}=3S_n$, then the ratio $S_{3n}/S_n$ is equal to

Answer 1

We know that

$S_n=\frac{n(n+1)}{2}$

Now, Given that

$S_{2n}=3S_n$

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

$\Rightarrow 2(2n+1)=3(n+1)$

$\Rightarrow 4n+2=3n+3$

$\Rightarrow n=1$

Therefore,

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

$\frac{S_{3n}}{S_n}=\frac{3n(3n+1)}{n(n+1)}$

Put $n=1$

$\frac{S_{3n}}{S_n}=\frac{3\times 4}{1\times 2}$

$\frac{S_{3n}}{S_n}=6$

Hence, this is the answer.