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

The Correct Option is A

The correct answer is A:\(\frac{4}{ac}-\frac{3}{b^2}\)
Given that:
\(\frac{1}{a},\frac{1}{b}\space and\space \frac{1}{c}\) are in A.P,
we know that, if the series is in A.P, then the difference of the consecutive terms are equal
i.e, \(\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\)  -(i)
\(\therefore\) as per the question we need to find:
\((\frac{1}{a}+\frac{1}{b}-\frac{1}{c})(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})\)
\(=(\frac{1}{a}+\frac{1}{b}-\frac{1}{a})(\frac{1}{c}+\frac{1}{c}-\frac{1}{6})\) [from equation (i)]
\(=(\frac{2}{a}-\frac{1}{b})(\frac{2}{c}-\frac{1}{b})\)
\(=\frac{4}{ac}-\frac{2}{ab}-\frac{2}{bc}+\frac{1}{b^2}\)
\(=\frac{4}{ac}-\frac{2}{b}(\frac{1}{a}+\frac{1}{c})+\frac{1}{b^2}\)
\(=\frac{4}{ac}-\frac{4}{b^2}+\frac{1}{b^2}\)    \((\therefore \frac{1}{a}+\frac{1}{c}=\frac{2}{b})\)
\(=\frac{4}{ac}-\frac{3}{b^2}\)