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If $\mathbf{S}_{\mathbf{n}}$ denote the sum of the first n terms of an A.P. If $S_{2 n}=3 S_{n}$ , then $\mathbf{S}_{3 \mathrm{n}}: \mathbf{S}_{\mathbf{n}}$ is equal to

A. 4 B. 6 C. 8 D. 10 Answer: Option B
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Solution(By Apex Team)

$\begin{aligned}S_n&=\text{ Sum of }n\text{ terms of A.P. and }S_{2n}=3S_n\\ S_n&=\frac{n}{2}[2a+(n-1)d]\\ S_{2n}&=\frac{2n}{2}[2a+(2n-1)d]\text{ and }\\ S_{3n}&=\frac{3n}{2}[2a+(3n-1)d]\\ &\text{We know that}\\ &S_{3n}=3\left(S_{2n}-S_n\right)\text{ and }S_{2n}=3S_n\\ &\Rightarrow\frac{S_{3n}}{S_n}=\frac{3\left(S_{2n}-S_n\right)}{S_n}\\ &\Rightarrow\frac{S_{3n}}{S_n}=\frac{3\left(3S_n-S_n\right)}{S_n}\\ &\Rightarrow\frac{S_{3n}}{S_n}=\frac{3\times2S_n}{S_n}\\ &\Rightarrow\frac{S_{3n}}{S_n}=\frac{6}{1}\\ &\therefore S_{3n}\ :\ S_n=6\end{aligned}$

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