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(D) Since $a, b, c$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$This implies $\frac{1}{b} - \frac{1}{a} = \fr

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none of these

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(D) Since $a, b, c$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$  .....$(1)$Now,consider the expression $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$.From $(1)$,we have $\frac{1}{c} - \frac{1}{a} = \frac{1}{b} - \frac{2}{b} = -\frac{1}{b}$ (not directly useful) or $\frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$.Substitute $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$ into the second bracket:$\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{2}{b} - \frac{1}{b} \right) = \left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{b} \right)$.From $(1)$,$\frac{1}{c} - \frac{1}{a} = \frac{2}{b} - \frac{2}{a}$. Substituting this:$\left( \frac{1}{b} + \frac{2}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.Alternatively,using $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$ in the first bracket:$\left( \frac{1}{b} + \frac{2}{b} - \frac{1}{a} - \frac{1}{a} \right) \left( \frac{1}{b} \right) = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.Since this result is not among the options,the correct answer is $D$.

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

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