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. \(\dfrac{4}{ac}-\dfrac{3}{b^2}\)
• B. \(\dfrac{b^2-ac}{a^2b^2c^2}\)
• C. \(\dfrac{4}{ac}-\dfrac{1}{b^2}\)
• D. None of these

Hint:

If reciprocals are in A.P., use the middle-term property directly.

Answer 1

The Correct Option is A

Step 1: Since \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) are in A.P.: \[ \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \]
Step 2: \[ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2}{b}+\frac{1}{b}=\frac{3}{b} \]
Step 3: \[ \left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)=\frac{2}{b}-\frac{2}{a} \]
Step 4: Multiplying and simplifying gives: \[ \frac{4}{ac}-\frac{3}{b^2} \]