If $\frac{a+b}{1-ab},b,\frac{b+c}{1-bc}$ are in A.P., then$\frac{1}{a},b,\frac{1}{c}$

$b-\frac{a+b}{1-ab}=\frac{b+c}{1-bc}-b$

$\Rightarrow \frac{b-a{b}^2-a-b}{1-ab}=\frac{b+c-b+b^{2}c}{1-bc}$

$\begin{array}{l}\Rightarrow -\frac{a\left(b^2+1\right)}{1-ab}=\frac{c\left(1+b^2\right)}{1-bc}\\ \Rightarrow -a+abc=c-abc\\ \Rightarrow 2abc=a+c\\ 2b=\frac{1}{a}+\frac{1}{c}\end{array}$

$\therefore \frac{1}{a},b,\frac{1}{c}$ are in A.P.