$\text{ Let }{S}_n\text{ denote the sum of the first n-terms of an A.P. If }{S}_{2\mathrm{n}}=3\cdot S_n\text{ then }{S}_{3n}÷S_n$

${\mathrm{S}}_{2\mathrm{n}}=3.{\mathrm{S}}_{\mathrm{n}}$

$\begin{array}{l}\Rightarrow \frac{2n}{2}[2a+(2n-1\left)d\right]=3\cdot \frac{n}{2}[2a+(n-1\left)d\right]\\ \Rightarrow 4a+2\left(2n-1\right)d=6a+3\left(n-1\right)d\\ \Rightarrow 2a=(n+1)d\\ \therefore \frac{s_{3n}}{s_n}=\frac{\frac{3n}{2}\left[2a+(3n-1)d\right]}{\frac{n[2a+(n-1)d}{2}}=\frac{3\left(4nd\right)}{2nd}=6\end{array}$