Thank you for reporting, we will resolve it shortly
Q.
If $a, b, c$ are in H.P., $b, c, d$ are in $G.P$. and $c, d, e$ are in
$A.P$., then the value of $e$ is
Sequences and Series
Solution:
$ a, b, c$ are in $H.P. \Rightarrow b=\frac{2 a c}{a+c} \,\,\,\,\,\, (i) $
$ b, c, d$ are in $G.P. \Rightarrow c^{2}=b d \,\,\,\,\,(ii)$
$c, d, e$ are in $A.P. \Rightarrow d=\frac{c+e}{2} \,\,\,\,\,\,(iii)$
From (i), $a b+b c=2 a c$
$\Rightarrow c=\frac{a b}{2 a-b} \,\,\,\,\,\,\,(iv)$
From (iii) and (iv),
$d=\frac{1}{2}\left[\frac{a b}{2 a-b}+e\right] \,\,\,\,\, (v)$
From (ii), (iv) and (v),
$\frac{a^{2} b^{2}}{(2 a-b)^{2}}=\frac{b}{2}\left[\frac{a b}{2 a-b}+e\right]$
$\Rightarrow e=\frac{a b^{2}}{(2 a-b)^{2}}$