Question

If $\frac{b+c-a}{a},\frac{c+a-b}{b}and\frac{a+b-c}{c}$ are in A.P. and $a+b+c\ne$ 0, then

• A. $b=\frac{ac}{a+c}$
• B. $b=\frac{2ac}{a+c}$
• C. $b=\frac{a+c}{2}$
• D. $b=\sqrt{ac}$

Answer 1

Answer: (B)

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

Given sequence is an AP.
$\therefore \frac{c+a-b}{b}-\frac{b+c-a}{a}=\frac{a+b-c}{c}-\frac{c+a-b}{b}$
$\Rightarrow \frac{ac+a^2-ab-(b^2+bc-ab)}{ab}=\frac{ab+b^2-bc-(c^2+ac-bc)}{bc}$
$\Rightarrow \frac{ac+a^2-ab-b^2-bc+ab}{ab}=\frac{ab+b^2-bc-c^2-ac+bc}{bc}$
$\Rightarrow \frac{ac+a^2-b^2-bc}{a}=\frac{ab+b^2-c^2-ac}{c}$
$\Rightarrow \frac{a^2-b^2+ac-bc}{a}=\frac{b^2-c^2+ab-ac}{c}$
$\Rightarrow \frac{(a+b)(a-b)+c(a-b)}{a}=\frac{(b+c)(b-c)+a(b-c)}{c}$
$\Rightarrow \frac{(a-b)(a+b+c)}{a}=\frac{(b-c)(a+b+c)}{c}$
$\Rightarrow \frac{a-b}{a}=\frac{b-c}{c}$
$\Rightarrow ac-bc=ab-ac$
$\Rightarrow 2ac=ab+bc$
$\Rightarrow 2ac=b(a+c)$
$\Rightarrow b=\frac{2ac}{a+c}$
Option B is correct.