If a, b, c are in GP then a+b, 2b, b+c are in

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none of these

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Solution

The correct option is C HP
Given $b^{2} = a c$
Let us consider,
$\frac{1}{a + b} + \frac{1}{b + c}$
$= \frac{2 b + c + a}{(a + b) (b + c)}$
$= \frac{2 b + c + a}{a b + b^{2} + a c + b c}$
$= \frac{2 b + c + a}{a b + 2 b^{2} + b c}$
$= \frac{2 b + c + a}{(b) (2 b + c + a)}$
$= \frac{1}{b}$
$∴ a + b, 2 b, b + c$ are in HP

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