Let $a, b, c, d$ and and $e$ be real numbers such that $a>b>0$and $c>0\text{. }$if $\frac{1}{a},\frac{1}{b},\frac{1}{c}$are in A.P., $b,c,d$ are in G.P. and $c, d, e$ are in A. P., then $\frac{a{b}^2}{(2a-b{)}^2}$ is equal to

Let $r$ be the common ratio of the G.P. $b, c, d$, then

$b=\frac{c}{r},c=c,d=cr$

As $c, d, e$, are in A.P. 

$e=2d-c=2cr-c=c(2r-1)$

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

$\Rightarrow \frac{1}{a}=\frac{2}{b}-\frac{1}{c}=\frac{2r}{c}-\frac{1}{c}=\frac{2r-1}{c}$

Also $\frac{2}{b}-\frac{1}{a}=\frac{1}{c}$

$\begin{array}{l}\Rightarrow \frac{2a-b}{ab}=\frac{1}{c}\\ \therefore \frac{a{b}^2}{(2a-b{)}^2}=\frac{a{b}^2}{(ab/c{)}^2}=\frac{c^2}{a}\\ =c(2r-1)=e\end{array}$