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nth Term of A.P

Let Sn deno...

Question

Let $S_{n}$ denote the sum of first $n$ terms of an $A P$ , If $S_{2 n} = 3 S_{n}$ then find the ratio $S_{3 n} / S_{n}$ .

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Solution

Let the first term of AP be a and common difference be d
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Given: $S_{2 n} = 3 S_{n}$
$S_{2 n} = \frac{2 n}{2} [2 a + (2 n - 1) d]$ ;
$3 S_{n} = \frac{3 n}{2} [2 a + (n - 1) d]$
$n [2 a + 2 n d - d] = \frac{3 n}{2} [2 a + n d - d]$
$\Rightarrow 4 a + 4 n d - 4 d = 6 a + 3 n d - 3 d$
$\Rightarrow 2 a - n d + d$
$\Rightarrow 2 a + d (1 - n) = 0$
$\Rightarrow 2 a = (n - 1) d$ ………… $(1)$
To find: ratio $S_{3 n} / S_{n}$
$\frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{6 a + 9 n d - 3 d}{2 a + (n d) - d}$
From equation $(1)$
$\frac{S_{3 n}}{S_{n}} = \frac{3 n d - 3 d + 9 n d - 3 d}{n d - d + n d - d} = \frac{6 d (2 n - 1)}{d (2 n - 1)} = 6$
$\Rightarrow \frac{S_{3 n}}{S_{n}} = 6$ .

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