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
Correct Answer: C
Solution :
[c] Let a, b and c are H.P. \[\therefore \,\frac{1}{a},\frac{1}{b},\frac{1}{c}\] are in A.P. \[\therefore \,\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\] \[\Rightarrow \frac{1}{c}=\frac{2}{b}-\frac{1}{a}\] Consider \[\left( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} \right)\left( \frac{1}{c}+\frac{1}{a}-\frac{1}{b} \right)\] \[=\left( \frac{1}{b}+\frac{2}{b}-\frac{1}{a}-\frac{1}{a} \right)\left( \frac{2}{b}-\frac{1}{b} \right)\] Using \[\frac{1}{a}+\frac{1}{c}=\frac{2}{b}=\left( \frac{3}{b}-\frac{2}{a} \right)\left( \frac{1}{b} \right)=\frac{3}{{{b}^{2}}}-\frac{2}{ab}\]
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