2020/001) a,b,c are in AP, b,c,d are in GP, and c,d,e are in HP, prove that a,c,e are in GP

Solution

We are given a,b,c are in AP so
$2b= a +c \cdots(1)$
b,c,d are in GP so
$bd=c^2\cdots(2)$
c,d,e are in HP, so
$\frac{1}{c} + \frac{1}{e} = \frac{2}{d}$
or $\frac{e+c}{ce} = \frac{2}{d}\cdots(3)$
or $d = \frac{2ce}{e+c}$
From(2)
$c^2 = bd$
or $2c^2 = 2bd = (a+c) \frac{2ce}{e+c}$ putting from (1) and (3)
or $c^2(e+c) = ce(a+c)$
or $c(e+c) = e(a+c)$
or $c^2= ae$

Hence  a,c,e are in GP