Let $S_{n}$ denote the sum of the first n terms of an AP. If $S_{2 n}$ =3 $S_{n}$ then $S_{3 n} : S_{n}$ is equal to

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Solution

The correct option is C 6
By using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ we have

Given

$\frac{2 n}{2} (2 a + (2 n - 1) d) = 3 \frac{n}{2} (2 a + (n - 1) d)$

$\Rightarrow 4 a + 4 n d - 2 d = 6 a + 3 n d - 3 d$

$\Rightarrow 2 a = (n + 1) d$

$S_{3 n} : S_{n} = \frac{\frac{3 n}{2} (2 a + (3 n - 1) d)}{\frac{n}{2} (2 a + (n - 1) d)}$

$= 6$

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