Question

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

Answer 1

Hint: Think of the basic definition of Harmonic progression, and use the property that the reciprocals of the terms in H.P. forms an A.P. We also know that the absolute value of difference of two consecutive terms of an A.P. is equal to the common difference of the Arithmetic progression, i.e., if three numbers x, y , z are in A.P. then y-x=z-y=common difference.

Complete step by-step answer:

Before starting with the solution to the above question, let us talk about harmonic progression.
In mathematics, harmonic progression is defined as a sequence of numbers, for which the sequence of the reciprocals of its terms gives an arithmetic progression.
We can represent a general harmonic progression as:$\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............$
Where $a,a+d,a+2d...............$ will represent an arithmetic progression.
Now let us start the solution to the above question. As it is given that a, b, c is in H.P., so we can say that $\dfrac{1}{a},\dfrac{1}{b}\text{ and }\dfrac{1}{c}$ are in A.P.
$\left( \dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a} \right)\left( \dfrac{1}{c}+\dfrac{1}{a}-\dfrac{1}{b} \right)$
Now we know that the difference of two consecutive terms of an A.P. is equal to the common difference of the Arithmetic progression. So, we let the common difference of the A.P. $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ be k.
  \[\left( \dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c} \right)\left( \dfrac{1}{c}-\left( \dfrac{1}{b}-\dfrac{1}{a} \right) \right)\]
\[=\left( k+\dfrac{1}{c} \right)\left( \dfrac{1}{c}-k \right)\]
Now, as we know that k is the common difference of A.P., the $\dfrac{1}{c}-k=\dfrac{1}{b}$ . Also, we can write $k=\dfrac{1}{c}-\dfrac{1}{b}$ .
\[\left( \dfrac{1}{c}-\dfrac{1}{b}+\dfrac{1}{c} \right)\dfrac{1}{b}\]
\[=\left( \dfrac{2}{c}-\dfrac{1}{b} \right)\dfrac{1}{b}\]
Therefore, the answer to the above question is \[\left( \dfrac{2}{c}-\dfrac{1}{b} \right)\dfrac{1}{b}\] .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is $\dfrac{1}{c}+\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{c}-\left( \dfrac{1}{a}-\dfrac{1}{b} \right)$ . Also, you need to remember the properties related to A.P. and H.P., as they are used very often.