Since 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. Then 2b = a + c ...(i) Also, $c^2$ = bd and $\frac{2}{d}$ = $\frac{1}{c}+\frac{1}{e}$ ⇒ $\frac{2ce}{d}$ = c + e ⇒ d = $\frac{2ce}{c+e}$ ... (iii) Multiplying (i) and (iii), we get 2bd = $\frac{2ce(a+c)}{c+e}$ ⇒ bd = $\frac{ce(a+c)}{c+e}$ ⇒ $c^2$ = $\frac{ce(a+c)}{c+e}$ [using (ii)] ⇒ c = $\frac{e(a+c)}{c+e}$ ⇒ c (c + e) = e (a + c) ⇒ $c^2$ + ce = ea + ce ⇒ $c^2$ = ae Hence above condition shows that a, c, e are in G.P.