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
Correct Answer: A
Solution :
\[\frac{1}{a},\frac{1}{b},\frac{1}{c}\]are in AP \[\therefore \] \[\frac{1}{a}+\frac{1}{c}=2/b\] ...(i) Hence, \[\left( \frac{1}{a}+\frac{1}{b}-\frac{1}{c} \right)\left( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} \right)\] \[=\left\{ \frac{1}{a}+\frac{1}{b}-\left( \frac{2}{b}-\frac{1}{a} \right) \right\}\left\{ \frac{1}{b}+\frac{1}{c}-\left( \frac{2}{b}-\frac{1}{c} \right) \right\}\] From Eq. (i) \[=\left( \frac{2}{a}-\frac{1}{b} \right)\left( \frac{2}{c}-\frac{1}{b} \right)=\frac{4}{ac}-\frac{2}{b}\left( \frac{1}{a}+\frac{1}{c} \right)+\frac{1}{{{b}^{2}}}\] \[=\frac{4}{ac}-\frac{2}{b}\left( \frac{2}{b} \right)+\frac{1}{{{b}^{2}}}=\frac{4}{ac}-\frac{3}{{{b}^{2}}}\]
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