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Question: If *a*, *b*, *c* are in H.P., then the value of \(\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \ri...

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

A

2bc+1b2\frac{2}{bc} + \frac{1}{b^{2}}

B

3c2+2ca\frac{3}{c^{2}} + \frac{2}{ca}

C

3b2−2ab\frac{3}{b^{2}} - \frac{2}{ab}

D

None of these

Answer

3b2−2ab\frac{3}{b^{2}} - \frac{2}{ab}

Explanation

Solution

a, b, c are in H.P. ⇒ 1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c} are in A.P. ∴ 1a+1c=2b\frac{1}{a} + \frac{1}{c} = \frac{2}{b}

Now, \left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right)\left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right) = \left{ \frac{1}{b} + \left( \frac{1}{a} + \frac{1}{c} \right) - \frac{2}{a} \right} (2b−1b)=(1b+2b−2a)(1b)=1b(3b−2a)=3b2−2ab\left( \frac{2}{b} - \frac{1}{b} \right) = \left( \frac{1}{b} + \frac{2}{b} - \frac{2}{a} \right)\left( \frac{1}{b} \right) = \frac{1}{b}\left( \frac{3}{b} - \frac{2}{a} \right) = \frac{3}{b^{2}} - \frac{2}{ab}