$S_{r}$ denotes the sum of the first r terms of an AP. Then $S_{3 n} : (S_{2 n} - S_{n})$ is

n

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3 n

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3

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independent of

n

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Solution

The correct options are
C $3$
D independent of $n$
The sum of an A.P with first term $a$ and common difference $d$ with $n$ terms is
$S_{n} = \frac{n}{2} [2 a + (n - 1)] d \Rightarrow S_{n} = \frac{n}{2} [2 a - d + n d]$
Similarly with 2n terms
$S_{2 n} = \frac{2 n}{2} [2 a + (2 n - 1) d] \Rightarrow S_{2 n} = \frac{2 n}{2} [2 a - d + 2 n d]$
And with 3n terms
$S_{3 n} = \frac{3 n}{2} [2 a + (3 n - 1) d] \Rightarrow S_{3 n} = \frac{3 n}{2} [2 a - d + 3 n d]$
$∴ S_{3 n} : (S_{2 n} - S_{n}) = \frac{3 [2 a - d + 3 n d]}{[2 a - d + 3 n d]} = 3$

Suggest Corrections

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If S_rdenotes the sum of the first r terms of an AP . Then find

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