Question

State True or False for the following statement. If a, b, and c are in A.P. then the following are also in A.P.:
$a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$

Answer 1

Hint: Perform the following operations and all the terms in order: divide by abc, multiply by ab + bc + ac, and subtract 1. Then simplify the terms and take a common from the first term, b common from the second term, and c common from the third term. Doing these, you will get the final answer.

Complete step by step answer:
In this question, we are given that a, b, and c are in A.P.
We need to find whether $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ are also in A.P.
We have, a, b, c are in A.P.
We know that if we perform the same operation on all the terms, then also they will remain in A.P.
So, dividing these by abc, we will get the following:
$\dfrac{a}{abc},\dfrac{b}{abc},\dfrac{c}{abc}$ are also in A.P.
Simplifying these terms, we will get the following:
$\dfrac{1}{bc},\dfrac{1}{ac},\dfrac{1}{ab}$ are also in A.P.
Now, multiply ab + bc + ac to all terms, we will get the following:
$\dfrac{ab+bc+ac}{bc},\dfrac{ab+bc+ac}{ac},\dfrac{ab+bc+ac}{ab}$ are in A.P.
Subtracting 1 from all the terms, we will get the following:
$\dfrac{ab+bc+ac}{bc}-1,\dfrac{ab+bc+ac}{ac}-1,\dfrac{ab+bc+ac}{ab}-1$ are in A.P.
Simplifying these terms, we will get the following:
$\dfrac{ab+ac}{bc},\dfrac{ab+bc}{ac},\dfrac{bc+ac}{ab}$ are also in A.P.
Now, taking a common from the first term, b common from the second term, and c common from the third term, we will get the following:
$\dfrac{a\left( b+c \right)}{bc},\dfrac{b\left( a+c \right)}{ac},\dfrac{c\left( a+b \right)}{ab}$ are also in A.P.
Simplifying these terms, we will get the following:
$a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ are also in A.P.
So, the given statement: If a, b, and c are in A.P., then the following are also in A.P.:$a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ is true.
This is our final answer.

Note: In this question, it is important to note that when three numbers are in A.P., and we perform the same operation on all the terms of this A.P., then the resulting terms will also be in A.P. This property is used throughout the solution and without this, you will not be able to solve the question.