Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A. P., then $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)$ is equal to

• A. $\frac{4}{ac}-\frac{3}{b^2}$
• B. $\frac{b^2-ac}{a^2{b}^2{c}^2}$
• C. $\frac{4}{ac}=\frac{1}{b^2}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{c}$
$\therefore \left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)$
$=\left(\frac{2}{a}-\frac{1}{b}\right)\left(\frac{2}{c}-\frac{1}{b}\right)=\frac{4}{ac}-\frac{1}{b}\left(\frac{2}{a}+\frac{2}{c}\right)+\frac{1}{b^2}$
$=\frac{4}{ac}-\frac{2}{b}\left(\frac{2}{b}\right)+\frac{1}{b^2}=\frac{4}{ac}-\frac{3}{b^2}$