Question

If $a,b,c,dande$ be five numbers such that $a,b,c$ are in $A.P.$, $b,c,d$ are in $G.P$., and $c,d,e$ are in $H.P.$, then prove that: $ae=(b-a{)}^2$.

Answer 1

We have,

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

So,

$2b=a+c$ $............\left(1\right)$

Since, $b,c,d$ are in $G.P$

So,

$c^2=bd$ $............\left(2\right)$

Since, $c,d,e$ are in $H.P$

So,

$\frac{1}{e}=\frac{2}{d}-\frac{1}{c}$ $..........\left(3\right)$

From equations $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

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

$\frac{1}{e}=\frac{2b}{(2b-a{)}^2}-\frac{1}{2b-a}$

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

$e=\frac{(2b-a{)}^2}{a}$

$ae=(2b-a{)}^2$

Hence, proved.