If $a, b, c$ are in $A.P.;$ $b, c, d$ are in $G.P.$ and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.$ prove that $a, c, e$ are in $G.P.$

It is given that $a, b, c$ are in $A.P.$

$\therefore b-a=c-b$       .......$(1)$

It is given that $b, c, d$ are in $G.P.$

$\therefore c^{2}=b d $       ........$(2)$

Also, $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.$

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\frac{2}{d}=\frac{1}{c}+\frac{1}{e}$       .........$(3)$

It has to be proved that $a, c, e$ are in $G.P.$ i.e., $c^{2}=a e$

From $(1),$ we obtain

$2 b=a+c$

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

From $(2),$ we obtain

$d=\frac{c^{2}}{b}$

Substituting these values in $(3),$ we obtain

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

$\Rightarrow \frac{2(a+c)}{2 c^{2}}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{a+c}{c^{2}}=\frac{e+c}{c e}$

$\Rightarrow \frac{a+c}{c}=\frac{e+c}{e}$

$\Rightarrow(a+c) e=(e+c) c$

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

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

Thus, $a, c$ and $e$ are in $G.P.$