Question

If $a,b,c$ are in $H.P$, $b,c,d$ are in $G.P$ and $c,d,e$ are in $A.P$ then $\frac{a{b}^2}{(2a-b{)}^2}$ is equal to

• A. $b$
• B. $a$
• C. $e$
• D. $d$

Answer 1

The correct option is C $e$

$a,b,c$ in H.P $⟹\frac{1}{a},\frac{1}{b},\frac{1}{c}$ in A.P

So Common difference $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

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

$⟹\frac{2a-b}{ab}=\frac{1}{c}⟹\frac{ab}{2a-b}=c$

$N=\frac{a{b}^2}{(2a-b{)}^2}=\frac{ab}{2a-b}\times \frac{b}{2a-b}=c\times \frac{c}{a}=\frac{c^2}{a}$

Since $b,c,d$ is G.P $⟹$ $c^2=bd$

Using $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

$⟹\frac{2}{b}-\frac{1}{c}=\frac{1}{a}⟹a=\frac{bc}{2c-b}$

$N=\frac{c^2}{a}=\frac{c^2}{(\frac{bc}{2c-b})}=\frac{c(2c-b)}{b}$

$N=\frac{2c^2-cb}{b}=\frac{2bd-cb}{b}=2d-c$

Since $c,d,e$ is given as A.P $⟹2d=c+e$

$N=2d-c=(c+e)-c=e$

Hence answer is option C