If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn , then S3n : Sn is equal to

Answer: Option B

$$S_n$$ = Sum of $$n$$ terms of A.P. and $$S_{2n}$$ = $$3S_n$$
$$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right],$$     $$S_{2n}=\frac{2n}{2}\left[2a+\left(2n-1\right)d\right]$$     and $$S_{3n}=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]$$
We know that $$S_{3n}=3\left(S_{2n}-S_n\right)$$    and $$S_{2n}=3S_n$$
$$\begin{array}{rl}& \Rightarrow \frac{S_{3n}}{S_n}=\frac{3\left(S_{2n}-S_n\right)}{S_n}\\ & \Rightarrow \frac{S_{3n}}{S_n}=\frac{3\left(3S_n-S_n\right)}{S_n}\\ & \Rightarrow \frac{S_{3n}}{S_n}=\frac{3\times 2S_n}{S_n}\\ & \Rightarrow \frac{S_{3n}}{S_n}=\frac{6}{1}\\ & \therefore S_{3n}:S_n=6\end{array}$$