If $\mathbf{S}_{\mathbf{n}}$ denotes the sum of the first r terms of an A.P. Then, $\mathbf{S}_{3 \mathrm{n}}:\left(\mathbf{S}_{2 \mathrm{n}}-\mathbf{S}_{\mathbf{n}}\right)$ is
A. n B. 3n C. 3 D. None of these Answer: Option CShow Answer
Solution(By Apex Team)
$\begin{aligned}S_n&=\frac{n}{2}[2a+(n-1)d]\\
S_{2n}&=\frac{2n}{2}[2a+(2n-1)d]\text{ and }\\
S_{3n}&=\frac{3n}{2}[2a+(3n-1)d]\\
&\text{ Now }S_{2n}-S_n&\\
&=\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d]\\
&=\frac{n}{2}[4a+(4n-2)d]-[2a+(n-1)d]\\
&=\frac{n}{2}[4a-2a+(4n-2-n+1)d]\\
&=\frac{n}{2}[2a+(3n-1)d]\\
&=\frac{1}{3}\left(S_{3n}\right)\\
&\therefore S_{3n}:\left(S_{2n}-S_n\right)\\
&=3:1\text{ or }\frac{3}{1}=3\end{aligned}$
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