Let S_{n} be the sum of the first n terms of | vedclass

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Question

Let a be first term and $d$ be common diff. of this A.P.

Given $\mathrm{S}_{3 \mathrm{n}}=3 \mathrm{~S}_{2 \mathrm{n}}$

$\Rightarrow \frac{3 n}{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

Let a be first term and $d$ be common diff. of this A.P.

Given $\mathrm{S}_{3 \mathrm{n}}=3 \mathrm{~S}_{2 \mathrm{n}}$

$\Rightarrow \frac{3 n}{2}[2 a+(3 n-1) d]=3 \frac{2 n}{2}[2 a+(2 n-1) d]$

$\Rightarrow 2 a+(3 n-1) d=4 a+(4 n-2) d$

$\Rightarrow 2 a+(n-1) d=0$

$	ext { Now } \frac{S_{a n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}=\frac{2[\underbrace{2 a+(n-1) d}_{-0}+3 n d]}{[\underbrace{2 a+(n-1) d}_{-0}+n d]}$

$=\frac{6 n d}{n d}=6$

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n}=3 S_{2 n}$, then the value of $\frac{S_{4 n}}{S_{2 n}}$ is:

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