Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $AP$, then $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)=$

• A. $(4b^2-3ac)/abc$
• B. $4/ac–3/b^2$
• C. $4/ac–5/b^2$
• D. $(4b^2+3ac)/a{b}^{2}c$

Answer 1

The correct option is B

$4/ac–3/b^2$

Finding the value of $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)$:

Given,

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $AP.$

$\Rightarrow$ $\frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{c}$ or $\frac{1}{c}+\frac{1}{a}=\frac{2}{b}$

Now,

$\begin{array}{rcl}\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)& =& \left(\frac{2}{a}-\frac{1}{b}\right)\left(\frac{2}{c}-\frac{1}{b}\right)\\ & =& \frac{4}{ac}-\frac{2}{bc}-\frac{2}{ab}+\frac{1}{b^2}\\ & =& \frac{4}{ac}-\left(\frac{1}{b}\right)\left[\frac{2}{c}+\frac{2}{a}\right]+\frac{1}{b^2}\\ & =& \frac{4}{ac}-\left(\frac{2}{b}\right)\left(\frac{2}{b}\right)+\frac{1}{b^2}\\ & =& \frac{4}{ac}-\left(\frac{4}{b^2}\right)+\frac{1}{b^2}\\ & \mathbf{=}& \frac{\mathbf{4}}{\mathbf{a}\mathbf{c}}\mathbf{-}\frac{\mathbf{3}}{{\mathbf{b}}^{\mathbf{2}}}\end{array}$

Hence, option(B) is correct.