If $\frac{1}{a + b}, \frac{1}{2 b}, \frac{1}{b + c}$ are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.

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Solution

Since, $\frac{1}{a + b}, \frac{1}{2 b}, \frac{1}{b + c}$ are in A.P.

$\frac{2}{2 b} = \frac{1}{(a + b)} \frac{1}{b + c}$

$\frac{1}{b} = \frac{b + c + a + b}{(a + b) (b + c)}$

$\frac{1}{b} = \frac{2 b + c + a}{a b + a c + b^{2} + b c}$

$a b + a c + b^{2} + b c = 2 b^{2} + b c + b a$

$b^{2} + a c = 2 b^{2}$

$b^{2} = a c$

So,

a, b, c are in G.P.

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