Question

If $a,b,c$ are in H. P., then the value of $(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$ is

• A. $\frac{2}{bc}-\frac{1}{b^2}$
• B. $\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$
• C. $\frac{3}{b^2}-\frac{1}{ab}$
• D. none of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{2}{bc}-\frac{1}{b^2}$
B $\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$
C $\frac{3}{b^2}-\frac{1}{ab}$

If a, b, c are in H.P., then $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\therefore$ Let E $=(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$

$\Rightarrow E=(\frac{3}{b}-\frac{2}{a})\left(\frac{1}{b}\right)$

$\therefore E=\frac{3}{b^2}-\frac{2}{ab}$ option C

Again, $(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$

$\Rightarrow E=(\frac{2}{c}-\frac{1}{b})\left(\frac{1}{b}\right)$

$\therefore E=\frac{2}{bc}-\frac{1}{b^2}$ option A

Also, $E=(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$

$\Rightarrow E=(\frac{1}{2}(\frac{1}{a}+\frac{1}{c})+\frac{1}{c}-\frac{1}{a})\left(\frac{1}{2}(\frac{1}{a}+\frac{1}{c})\right)$

$E=\frac{1}{4}{(\frac{1}{a}+\frac{1}{c})}^2+\frac{1}{2}(\frac{1}{c^2}-\frac{1}{a^2})$

$\therefore E=\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$ option B