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.

Answer 1

Given that $a,b,c$ are in A.P.

$\Rightarrow b-a=c-b$

$\Rightarrow b=\frac{a+c}{2}$ ...(1)

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

$\Rightarrow c^2=bd$ ...(2)

Also, given $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ 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)

We have to prove $a,c,e$ are in G.P.

i.e. $c^2=ae$

Substituting the value of $d$ from eqn (2) in eqn (3), we get

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

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

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

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

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

$\Rightarrow ae+ce=ec+c^2$

$\Rightarrow c^2=ae$

Hence, $a,c,e$ are in G.P