Question

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

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

Answer: (B)

$6$

$S_{2n}=3S_n$
$\frac{2n}{2}(2a+(2n-1\left)d\right)=3\frac{n}{2}(2a+(n-1\left)d\right)$
$4a+4nd-2d=6a+3nd-3d$
nd +d = 2a ....(i)
$\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}(2a+(3n-1\left)d\right)}{\frac{n}{2}(2a+(n-1\left)d\right)}$
= $\frac{\frac{3n}{2}(nd+d+(3n-1\left)d\right)}{\frac{n}{2}(nd+d+(n-1\left)d\right)}$
= $\frac{\frac{3n}{2}\left(4nd\right)}{\frac{n}{2}\left(2nd\right)}$
= 6