Question

Let $S_n$ denote the sum of the first 'n' terms of an A.P. and $S_{2n}=3S_n$. Then, the ratio $S_{3n}:S_n$ is equal to

• A. $4:1$
• B. $6:1$
• C. $8:1$
• D. $10:1$

Answer 1

Answer: (B)

$6:1$

$S_n=\frac{n(n+1)}{2}$
now it is given that
$S_{2n}=3S_n$
$\Rightarrow \frac{2n(2n+1)}{2}=\frac{3n(n+1)}{2}$
$\Rightarrow 2(2n+1)=3(n+1)$
$\Rightarrow 4n+2=3n+3$
$\Rightarrow n=1$
Then $\frac{S_{3}n}{S_n}=\frac{\left[3n\right(3n+1\left)/2\right]}{\left[n\right(n+1\left)/2\right]}=\frac{3\times 4}{1\times 2}=6:1$