If the sum of the first $n$ terms of an arithmetic progression is $S_n$, and $S_{2n}=3S_n$, then find the ratio $\frac{S_{3n}}{S_n}$.

Solution

1. The sum of the first $n$ terms is given by $S_n=\frac{n}{2}[2a+(n-1)d]$.
2. Given $S_{2n}=3S_n$, we substitute the formula to get $\frac{2n}{2}[2a+(2n-1)d]=3\cdot \frac{n}{2}[2a+(n-1)d]$, which simplifies to $4a + 2(2n-1)d = 6a + 3(n-1)d$.
3. Rearranging the equation yields $2a = (n+1)d$.
4. The ratio $$\begin{array}{rl}\frac{S_{3n}}{S_n}& =\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\\ & =3\cdot \frac{(n+1)d+(3n-1)d}{(n+1)d+(n-1)d}\\ & =3\cdot \frac{4nd}{2nd}\\ & =6\end{array}$$.
5. Hence the answer is (C).