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

Solution

1. The sum is $S_n=\frac{n}{2}(2a+(n-1)d)$.
2. Given $S_{2n}=3S_n$, we have $\frac{2n}{2}(2a+(2n-1)d)=3\frac{n}{2}(2a+(n-1)d)$.
3. Simplifying gives $4a+2(2n-1)d=3a+\frac{3}{2}(n-1)d$, which leads to $a=-d(\frac{n+1}{2})$.
4. Substituting $a$ into $S_{3n}=\frac{3n}{2}(2a+(3n-1)d)$ yields $S_{3n}=6S_n$.
5. Hence the answer is (B).