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# If a, b, and c are in H.P., then the value of  $\frac{\left(\mathrm{ac}+\mathrm{ab}-\mathrm{bc}\right)\left(\mathrm{ab}+\mathrm{bc}-\mathrm{ac}\right)}{\left(\mathrm{abc}{\right)}^{2}}$ is

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a
$\frac{\left(\mathrm{a}+\mathrm{c}\right)\left(3\mathrm{a}-\mathrm{c}\right)}{4{\mathrm{a}}^{2}{\mathrm{c}}^{2}}$
b
$\frac{2}{\mathrm{bc}}-\frac{1}{{\mathrm{b}}^{2}}$
c
$\frac{2}{\mathrm{bc}}-\frac{1}{{\mathrm{a}}^{2}}$
d
$\frac{\left(\mathrm{a}-\mathrm{c}\right)\left(3\mathrm{a}+\mathrm{c}\right)}{4{\mathrm{a}}^{2}{\mathrm{c}}^{2}}$

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detailed solution

Correct option is A

$\begin{array}{l}\left(\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)\left(\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right)=\left(\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}-\frac{2}{\mathrm{b}}+\frac{1}{\mathrm{c}}\right)\left(\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{b}}-\frac{1}{\mathrm{c}}\right)\\ =\left(\frac{2}{\mathrm{c}}-\frac{1}{\mathrm{b}}\right)\frac{1}{\mathrm{b}}=\frac{2}{\mathrm{bc}}-\frac{1}{{\mathrm{b}}^{2}}\end{array}$
Also by eliminating b, we get the given expression
$\frac{\left(\mathrm{a}+\mathrm{c}\right)\left(3\mathrm{a}-\mathrm{c}\right)}{4{\mathrm{a}}^{2}{\mathrm{c}}^{2}}$