If a, b, c are in GP, prove that

Solution:

To prove: $\frac{1}{(a+b)}, \frac{1}{(2 b)}, \frac{1}{(b+c)}$ are in AP

Given: $a, b, c$ are in GP

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

When $a, b, c$ are in $G P, b^{2}=a c$

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

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

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

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

$\Rightarrow \frac{a+c+2 b}{a b+b^{2}+b^{2}+b c}\left[b^{2}=a c\right]$

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

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

$\Rightarrow \frac{1}{b}$

$\Rightarrow 2 \times \frac{1}{2 b}$

We can see that $\frac{1}{(a+b)}+\frac{1}{(b+c)}=2 \times \frac{1}{2 b}$

Hence we can say that $\frac{1}{(a+b)}, \frac{1}{(2 b)}, \frac{1}{(b+c)}$ are in AP.