Mark the correct alternative in the following question:

Let S_n denote the sum of first n terms of an A.P. If S₂_n = 3S_n, then S₃_n : S_n is equal to

(a) 4 (b) 6 (c) 8 (d) 10

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Solution

$As, S_{2 n} = 3 S_{n} \Rightarrow \frac{2 n}{2} [2 a + (2 n - 1) d] = \frac{3 n}{2} [2 a + (n - 1) d] \Rightarrow 2 [2 a + (2 n - 1) d] = 3 [2 a + (n - 1) d] \Rightarrow 4 a + 2 (2 n - 1) d = 6 a + 3 (n - 1) d \Rightarrow 4 a + 4 n d - 2 d = 6 a + 3 n d - 3 d \Rightarrow 6 a - 4 a + 3 n d - 3 d - 4 n d + 2 d = 0 \Rightarrow 2 a - n d - d = 0 \Rightarrow 2 a - (n + 1) d = 0 \Rightarrow 2 a = (n + 1) d . . . . . (i) Now, \frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{3 [(n + 1) d + (3 n - 1) d]}{[(n + 1) d + (n - 1) d]} [Using (i)] = \frac{3 [n d + d + 3 n d - d]}{[n d + d + n d - d]} = \frac{3 [4 n d]}{[2 n d]} = 3 \times 2 = 6$

Hence, the correct alternative is option (b).

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