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 :
\[a,\ b,\ c\]are in H.P., then \[\frac{1}{a},\ \frac{1}{b},\ \frac{1}{c}\] are in A.P. \[\Rightarrow \]\[\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\] Now, \[\left( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} \right)\,\left( \frac{1}{c}+\frac{1}{a}-\frac{1}{b} \right)\] \[=\left( \frac{3}{b}-\frac{2}{a} \right)\,\left( \frac{1}{b} \right)=\frac{3}{{{b}^{2}}}-\frac{2}{ab}\].
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