Mathematics - Sequence & Series Question with Solution | TestHub

Let the first term of the A.P. be $a$ and the common difference be $d$. The sum of the first $k$ terms is $S_k=\frac{k}{2}(2a+(k-1)d)$. Given $S_{2n}=3S_n$: $$\frac{2n}{2}(2a+(2n-1)d)=3\cdot \frac{n}{2}(2a+(n-1)d)$$ $$2(2a+(2n-1)d)=3(2a+(n-1)d)$$ $$4a+(4n-2)d=6a+(3n-3)d$$ $$2a=(n+1)d$$ Now, we need to find the ratio $S_{4n}:S_n$:$$S_{4n}=\frac{4n}{2}(2a+(4n-1)d)=2n((n+1)d+(4n-1)d)=2n(5n)d=10n^{2}d$$ $$S_n=\frac{n}{2}(2a+(n-1)d)=\frac{n}{2}((n+1)d+(n-1)d)=\frac{n}{2}(2n)d=n^{2}d$$ Therefore, $S_{4n}:S_n=10n^{2}d:n^{2}d=10:1$. The ratio is $10$.