Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 \mathrm{n}}=3 S_{2 \mathrm{n}}$, then the value
of $\frac{S_{4 n}}{S_{2 n}}$ is :
Correct Option: 1
Let a be first term and d be common diff. of this A.P.
Given $S_{3 \mathrm{n}}=3 S_{2 \mathrm{n}}$
$\Rightarrow \frac{3 \mathrm{n}}{2}[2 \mathrm{a}+(3 \mathrm{n}-1) \mathrm{d}]=3 \frac{2 \mathrm{n}}{2}[2 \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 2 \mathrm{a}+(3 \mathrm{n}-1) \mathrm{d}=4 \mathrm{a}+(4 \mathrm{n}-2) \mathrm{d}$
$\Rightarrow 2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=0$
Now $\frac{S_{4 n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}=\frac{2[\underbrace{2 a+(n-1) d}_{-0}+3 n d]}{[\underbrace{2 a+(n-1) d}_{-0}+n d]}$
$=\frac{6 \text { nd }}{\text { nd }}=6$