Question

$a,b,c\in R-\left\{0\right\}$ are in $H.P$. Then $\frac{b+a}{b-a}+\frac{b+c}{b-c}$ is equal to

Answer 1

We have,

$a,b,cinH.P.$

So,

$\frac{1}{a},\frac{1}{b},\frac{1}{c}......\text{in}\text{an}\text{A}\text{.P}$ $\therefore H.P.=\frac{1}{A.P.}$

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

$\Rightarrow \frac{1}{b}=\frac{a+c}{2ac}$

$\Rightarrow b=\frac{2ac}{a+c}$

Now,

$\frac{b+a}{b-a}+\frac{b+c}{b-c}$

$\Rightarrow \frac{(b+a)(b-c)+(b+c)(b-a)}{(b-a)(b-c)}$

$\Rightarrow \frac{b^2+ab-bc-ac+b^2+bc-ab-ac}{(b-a)(b-c)}$

$\Rightarrow \frac{2b^2-2ac}{(b-a)(b-c)}$

$\Rightarrow \frac{2\frac{2ac}{a+c}-2ac}{(\frac{2ac}{a+c}-a)(\frac{2ac}{a+c}-b)}$

$\Rightarrow \frac{\frac{4ac-2a^{2}c-2a{c}^2}{a+c}}{\left(\frac{2ac-a^2-ac}{a+c}\right)\left(\frac{2ac-ab-bc}{a+c}\right)}$

$\Rightarrow \frac{\frac{4ac-2a^{2}c-2a{c}^2}{a+c}}{\frac{(2ac-a^2-ac)(2ac-ab-bc)}{a+c}}$

$\Rightarrow \frac{4ac-2a^{2}c-2a{c}^2}{(ac-a^2)(2ac-ab-bc)}$

$\Rightarrow \frac{ac(4-2a-2c)}{a(c-a)(2ac-ab-bc)}$

$\Rightarrow \frac{c(4-2a-2c)}{(c-a)(2ac-ab-bc)}$

Hence, this is the answer.