Question
If $$a, b$$ and $$c$$ are in H. P. then the value of $$\left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right)\left( {\frac{1}{c} + \frac{1}{a} - \frac{1}{b}} \right){\text{is :}}$$
A.
$$\frac{2}{{bc}} + \frac{1}{{{b^2}}}$$
B.
$$\frac{3}{{c^2}} + \frac{2}{{{ca}}}$$
C.
$$\frac{3}{{b^2}} - \frac{2}{{{ab}}}$$
D.
None of these
Answer :
$$\frac{3}{{b^2}} - \frac{2}{{{ab}}}$$
Solution :
Let $$a, b$$ and $$c$$ are in H.P.
$$\eqalign{
& \therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr
& \therefore \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \cr
& \Rightarrow \frac{1}{c} = \frac{2}{b} - \frac{1}{a} \cr
& {\text{Consider }}\left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right)\left( {\frac{1}{c} + \frac{1}{a} - \frac{1}{b}} \right) \cr
& = \left( {\frac{1}{b} + \frac{2}{b} - \frac{1}{a} - \frac{1}{a}} \right)\left( {\frac{2}{b} - \frac{1}{b}} \right) \cr
& {\text{Using }}\frac{1}{a} + \frac{1}{c} = \frac{2}{b} = \left( {\frac{3}{b} - \frac{2}{a}} \right)\left( {\frac{1}{b}} \right) = \frac{3}{{{b^2}}} - \frac{2}{{ab}} \cr} $$