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Arithmetic Progression

Consider an A...

Question

Consider an A, P with first term $^{'} a^{'}$ and the common difference $d$ . Let $S_{k}$ denote the sum of the first $K$ terms. Let $\frac{S_{k x}}{S_{x}}$ is independent of $x$ , then -

a = \frac{d}{2}

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a = d

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a = 2 d

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None of these

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Solution

The correct option is A $a = \frac{d}{2}$
$\frac{S_{k x}}{S_{x}} = \frac{\frac{k x}{2} [2 a + (k x - 1) d]}{\frac{x}{2} [2 a + (x - 1) d]} = \frac{k [2 a + (k x - 1) d]}{[2 a + (x - 1) d]}$
$= \frac{k [(2 a - d) + k x d]}{[(2 a - d) + x d]}$
Now , $\frac{S_{k x}}{S_{x}}$ is independent of $x$ if $2 a - d = 0$

$\Rightarrow a = \frac{d}{2}$

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