Question

If $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in AP, then find the value $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right)$.
A. $\dfrac{{4{b^2} - 3ac}}{{abc}}$
B. $\dfrac{4}{{ac}} - \dfrac{3}{{{b^2}}}$
C. $\dfrac{4}{{ac}} - \dfrac{5}{{{b^2}}}$
D. $\dfrac{{4{b^2} + 3ac}}{{a{b^2}c}}$

Answer 1

Hint: First we will apply the condition of AP to the series $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$. From the condition, we will find the value of $\dfrac{1}{a}$. Then substitute the value of $\dfrac{1}{a}$ and $\dfrac{1}{c}$ in the given expression and simplify the expression to get the desired result.

Formula Used:
If $a,b,c$ are in AP, then $b - a = c - b$.

Complete step by step solution:
Given that, $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in AP.
Now apply the condition of AP.
$\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{1}{b}$
Now subtract $\dfrac{1}{b}$ from both sides.
$ \Rightarrow \dfrac{1}{b} - \dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{c} - \dfrac{1}{b} - \dfrac{1}{b}$
$ \Rightarrow - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{2}{b}$
Multiply -1 both sides of the equation.
$ \Rightarrow \dfrac{1}{a} = \dfrac{2}{b} - \dfrac{1}{c}$
$ \Rightarrow \dfrac{2}{b} - \dfrac{1}{a} = \dfrac{1}{c}$
Now substitute the value of $\dfrac{1}{a}$ in $\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right)$ and $\dfrac{1}{c}$ in $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)$.
$\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right)$
$\Rightarrow \left( {\dfrac{1}{a} + \dfrac{1}{b} - \left( {\dfrac{2}{b} - \dfrac{1}{a}} \right)} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \left( {\dfrac{2}{b} - \dfrac{1}{c}} \right)} \right)$
Now add or subtract all like terms.
$\Rightarrow \left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{2}{b} + \dfrac{1}{a}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{2}{b} + \dfrac{1}{c}} \right)$
$\Rightarrow \left( {\dfrac{2}{a} - \dfrac{1}{b}} \right)\left( {\dfrac{2}{c} - \dfrac{1}{b}} \right)$
Multiply in both terms
$\Rightarrow \dfrac{4}{{ac}} - \dfrac{2}{{bc}} - \dfrac{2}{{ab}} + \dfrac{1}{{{b^2}}}$
$\Rightarrow \dfrac{4}{{ac}} + \dfrac{1}{{{b^2}}} - \dfrac{2}{{bc}} - \dfrac{2}{{ab}}$
Now take common $ - \dfrac{2}{b}$ from the last two terms
$\Rightarrow \dfrac{4}{{ac}} + \dfrac{1}{{{b^2}}} - \dfrac{2}{b}\left( {\dfrac{1}{c} + \dfrac{1}{a}} \right)$
$\Rightarrow \dfrac{4}{{ac}} + \dfrac{1}{{{b^2}}} - \dfrac{2}{b}\left( {\dfrac{2}{b}} \right)$ Since $ - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{2}{b} \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
$\Rightarrow \dfrac{4}{{ac}} + \dfrac{1}{{{b^2}}} - \dfrac{4}{{{b^2}}}$
Subtract like terms
$\Rightarrow \dfrac{4}{{ac}} - \dfrac{3}{{{b^2}}}$

Option ‘B’ is correct

Note: To calculate the value of the given expression, we should put the value of $\dfrac{1}{a}$ in $\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right)$ and $\dfrac{1}{c}$ in $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)$.