Solve the following


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.


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

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

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

$\Rightarrow(a+b)(b+c)=b(2 b+a+c)$

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

$\Rightarrow 2 b^{2}-b^{2}=a c$

$\Rightarrow b^{2}=a c$

Thus, $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are in G.P.

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