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 $G.P.$

(N/A) It is given that $a, b, c$ are in $A.P.$
$\therefore 2b = a + c$ .......$(1)$
It is given that $b, c, d$ are in $G.P.$
$\therefore c^{2} = bd$ .......$(2)$
Also,$\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.$
$\therefore \frac{2}{d} = \frac{1}{c} + \frac{1}{e}$ .......$(3)$
We need to prove that $a, c, e$ are in $G.P.,$ i.e.,$c^{2} = ae.$
From $(1),$ $b = \frac{a+c}{2}.$
From $(2),$ $d = \frac{c^{2}}{b}.$
Substituting these values into $(3):$
$\frac{2}{\frac{c^{2}}{b}} = \frac{1}{c} + \frac{1}{e}$
$\frac{2b}{c^{2}} = \frac{e+c}{ce}$
$\frac{2(\frac{a+c}{2})}{c^{2}} = \frac{e+c}{ce}$
$\frac{a+c}{c^{2}} = \frac{e+c}{ce}$
$\frac{a+c}{c} = \frac{e+c}{e}$
$(a+c)e = c(e+c)$
$ae + ce = ce + c^{2}$
$c^{2} = ae$
Thus,$a, c, e$ are in $G.P.$