Question

If $\frac{1}{a},\frac{1}{H},\frac{1}{b}$are in $AP$, then $\frac{(H+a)}{(H-a)}+\frac{(H+b)}{(H-b)}=?$

• A. $2$
• B. $4$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$2$

.Explanation for the correct option:

Given,

$\frac{1}{a},\frac{1}{H},\frac{1}{b}$ are in $AP$

$\frac{2}{H}=\frac{1}{a}+\frac{1}{b}$

$\Rightarrow$ $\frac{2}{H}=\frac{\left(b+a\right)}{ab}$

$\Rightarrow$ $\frac{H}{2}=\frac{ab}{\left(a+b\right)}$

$\Rightarrow$ $H=\frac{2ab}{\left(a+b\right)}$

$\Rightarrow$ $\left(H+a\right)=\left[\frac{2ab}{\left(a+b\right)}\right]+a$

$=\left[\frac{2ab+a\left(a+b\right)}{\left(a+b\right)}\right]$

$=\left[\frac{a\left(2b+a+b\right)}{\left(a+b\right)}\right]$

$\Rightarrow$ $\left(H+a\right)$ $=\left[\frac{a\left(a+3b\right)}{\left(a+b\right)}\right]$

Similarly,

$\left(H-a\right)=\left[\frac{2ab}{\left(a+b\right)}\right]-a$

$=\left[\frac{2ab-a\left(a+b\right)}{\left(a+b\right)}\right]$

$=\left[\frac{a\left(2b-a-b\right)}{\left(a+b\right)}\right]$

$\Rightarrow$$\left(H-a\right)$$=\left[\frac{a\left(b-a\right)}{\left(a+b\right)}\right]$

$\frac{\left(H+a\right)}{\left(H-a\right)}=\frac{\left[{\displaystyle \frac{a\left(a+3b\right)}{\left(a+b\right)}}\right]}{\left[{\displaystyle \frac{a\left(b-a\right)}{\left(a+b\right)}}\right]}$

$=\frac{a\left(a+3b\right)}{a\left(b-a\right)}$

$\Rightarrow$$\frac{H+a}{H-a}=\frac{\left(a+3b\right)}{\left(b-a\right)}\to \left(1\right)$

Similarly,

$\frac{\left(H+b\right)}{\left(H-b\right)}=\frac{\left(3a+b\right)}{\left(a-b\right)}\to \left(2\right)$

Add $\left(1\right)$ and $\left(2\right)$

$\frac{\left(H+a\right)}{\left(H-a\right)}+\frac{\left(H+b\right)}{\left(H-b\right)}=\left[\frac{\left(a+3b\right)}{\left(b-a\right)}\right]+\left[\frac{\left(3a+b\right)}{\left(a-b\right)}\right]$

$=\left[\frac{\left(a+3b\right)}{\left(b-a\right)}\right]+\left[\frac{\left(-3a-b\right)}{\left(b-a\right)}\right]$ (multiply and divided by negative on RHS $2nd$ equation)

$=\left[\frac{a+3b-3a-b}{b-a}\right]$

$=\frac{\left(2b-2a\right)}{\left(b-a\right)}$

$=\frac{2\left(b-a\right)}{\left(b-a\right)}$

$=2$

Hence, option(A) is correct.