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

a, b, c are in A.P.

$\therefore b-a=c-b\Rightarrow 2b=a+c$

$\Rightarrow b=\frac{a+c}{2}\dots \dots \left(1\right)$

$\because$ b, c, d are in G.P.

$\therefore \frac{c}{b}=\frac{d}{c}\Rightarrow c^2=bd\dots \dots \left(2\right)$

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

$\therefore \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}\Rightarrow \frac{2}{d}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2}{d}=\frac{c+e}{ce}\Rightarrow d=\frac{2ce}{c+e}\dots \dots \left(3\right)$

Putting value of equation (1) and (3) in (2)

$c^2=\left(\frac{c+a}{2}\right)\left(\frac{2ce}{c+e}\right)=\frac{ce[c+a]}{c+e}$

$\Rightarrow c^2[c+e]=ec[c+a]$

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

$\Rightarrow c^2=ae$

which shows that a, c, e are in G.P.