Let a, b, c, be in A.P. with a common difference d Then $e^{1/c},e^{b/ac},e^{1/a}$ are in:

$\mathrm{a},\mathrm{b},\mathrm{c}\text{ are in A.P. }\Rightarrow 2\mathrm{b}=\mathrm{a}+\mathrm{c}$

Now 

$e^{1/c}\times e^{1/a}=e^{(a+c)/ac}=e^{2b/ac}={\left(e^{b/ac}\right)}^2$

$\therefore {\mathrm{e}}^{1/\mathrm{c}},{\mathrm{e}}^{\mathrm{b}/\mathrm{ac}},{\mathrm{e}}^{1/\mathrm{a}}$ in GP. with common ratio

$\begin{array}{l}=\frac{e^{b/ac}}{e^{1/c}}=e^{(b-a)/ac}=e^{d/(b-d)(b+d)}\\ =e^{d/\left(b^2-d^2\right)}\end{array}$

$[\because$a b c are in A.P. with common difference $\mathrm{d}, \therefore \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b}=\mathrm{d}]$