Question

If $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP, then $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right) =$
(1) $\dfrac{{\left( {4{b^2} - 3ac} \right)}}{{abc}}$
(2) $\dfrac{4}{{ac}} - \dfrac{3}{{{b^2}}}$
(3) $\dfrac{4}{{ac}} - \dfrac{5}{{{b^2}}}$
(4) $\dfrac{{\left( {4{b^2} + 3ac} \right)}}{{a{b^2}c}}$

Answer 1

A sequence of numbers is called an arithmetic progression, if the difference between any two consecutive terms is always the same i.e., if $x,y,z$ are in AP then $y - x = z - y$ . In the question $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP. So, first we will just convert the given terms like the general condition of AP and then simplify it. Hence, we will get the required answer.

Complete step by step answer:
We are given that $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP
And we know that when variables are in AP, then this means that the difference between any two consecutive terms is always the same.
$\therefore {\text{ }}\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{1}{b}{\text{ }} - - - \left( 1 \right)$
Now we have to find the value of $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right)$
So, let us consider $\left( {\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a}} \right) - - - \left( A \right)$
From equation $\left( 1 \right)$ by shifting the terms we can write,
${\text{ }}\dfrac{1}{b} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{b}{\text{ }} - - - \left( 2 \right)$
Now substitute the values from equation $\left( 1 \right)$ and equation $\left( 2 \right)$ in equation $\left( A \right)$ we get
$\left( {\dfrac{1}{a} + \dfrac{1}{a} - \dfrac{1}{b}} \right)\left( {\dfrac{1}{c} - \dfrac{1}{b} + \dfrac{1}{c}} \right)$
$\Rightarrow \left( {\dfrac{2}{a} - \dfrac{1}{b}} \right)\left( {\dfrac{2}{c} - \dfrac{1}{b}} \right)$
On multiplying, we get
$\dfrac{4}{{ac}} - \dfrac{2}{{ab}} - \dfrac{2}{{bc}} + \dfrac{1}{{{b^2}}}$
Taking common $\dfrac{2}{b}$ from middle two terms, we get
$\dfrac{4}{{ac}} - \dfrac{2}{b}\left( {\dfrac{1}{a} + \dfrac{1}{c}} \right) + \dfrac{1}{{{b^2}}}{\text{ }} - - - \left( 3 \right)$
Now if we see from equation $\left( 1 \right)$ we can write,
$\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$
$\therefore$ equation $\left( 3 \right)$ becomes,
$\dfrac{4}{{ac}} - \dfrac{2}{b}\left( {\dfrac{2}{b}} \right) + \dfrac{1}{{{b^2}}}$
$\Rightarrow \dfrac{4}{{ac}} - \dfrac{4}{{{b^2}}} + \dfrac{1}{{{b^2}}}$
On simplifying it, we get
$\dfrac{4}{{ac}} - \dfrac{3}{{{b^2}}}$
which is the required answer.

So, the correct answer is “Option 2”.

Note:
One of the main things to note while solving these types of questions is that sometimes another formula can also be used i.e., if $x,y,z$ are in AP then $2y = x + z$ . So, don't get confused between the two. It also has the similar meaning which we have used to solve the question. It is just that sometimes the question is solved with the help of this formula, so we use this formula.
Since $2y = x + z$ can also be written as
$y + y = x + z$
$\Rightarrow y - x = z - y$ which is what we used in the problem.