If Sn denotes the sum of the first r terms of an A.P. Then, S3n : (S2n – Sn) is

Answer: Option C

$$\begin{array}{rl}& S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right],\\ & S_{2n}=\frac{2n}{2}\left[2a+\left(2n-1\right)d\right]\text{and}\\ & S_{3n}=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]\\ & \text{Now}{S}_{2n}-S_n\end{array}$$
$$=\frac{2n}{2}\left[2a+\left(2n-1\right)d\right]-$$     $$\frac{n}{2}\left[2a+\left(n-1\right)d\right]$$
$$=\frac{n}{2}\left[4a+\left(4n-2\right)d\right]-$$     $$\left[2a+\left(n-1\right)d\right]$$
$$\begin{array}{rl}& =\frac{n}{2}\left[4a-2a+\left(4n-2-n+1\right)d\right]\\ & =\frac{n}{2}\left[2a+\left(3n-1\right)d\right]\\ & =\frac{1}{3}\left(S_{3n}\right)\\ & \therefore S_{3n}:\left(S_{2n}-S_n\right)\\ & =3:1\text{or}\frac{3}{1}=3\end{array}$$