Question

If a,b,c,d,e 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
$e=\frac{(b-a{)}^2}{a}$

• A. True
• B. False

Answer 1

The correct option is B False

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

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

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

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

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

Hence, this is the answer.