 # If a, b, c are in A.P,; b, c, d are in G.P and are in A.P. prove that a, c, e are in G.P. `
Question:

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.

Solution:

It is aiven that $a \cdot b \cdot c$ are in A.P.

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

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

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

Also, $\frac{1}{c}, \frac{1}{d}, \frac{1}{c}$ 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, ac, and e are in G.P.