Let S_n be the sum of the first n terms of an arithmetic progression.

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 $\frac{S_{4 n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}=\frac{2[\underbrace{2 a+(n-1) d}_{-0}+3 n d]}{[\underbrace{2 a+(n-1) d}_{-0}+n d]}$

$=\frac{6 \text { nd }}{\text { nd }}=6$