Question

If $a,b,c$ are in AP and $b,c,d$ are in GP and $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in AP, prove that $a,c,e$ are in GP.

Answer 1

$a,b,c$ are in AP.

$b-a=c-b$ .......(1)

$b,c,d$ are in GP.

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

Also, $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in AP.

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\frac{2}{d}=\frac{1}{c}+\frac{1}{e}$ .......(3)

To prove : $a,c,e$ are in GP $⟹c^2=ae$.

From 1,

$b=\frac{a+c}{2}$

From 2, $d=\frac{c^2}{b}$

Substituting these values in 3, we get,

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

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

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

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

$ae+ce=ec+c^2$

$c^2=ae$

Thus, $a,c,e$ are in GP.