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

Question

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3n}=3S_{2n}$, then the value of $\dfrac{S_{4n}}{S_{2n}}$ is:

Accepted Answer

Write $S_n = n(An+B)$. From $S_{3n}=3S_{2n}$: $3n(3An+B)=3\cdot2n(2An+B) \Rightarrow 3An+B=4An+2B \Rightarrow An+B=0$.

$$\frac{S_{4n}}{S_{2n}} = \frac{4n(4An+B)}{2n(2An+B)} = 2\cdot\frac{4An-An}{2An-An} = 2\times3 = 6.$$

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