Question

If $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in AP, prove that $\dfrac{{\left( {b + c} \right)}}{a}$, $\dfrac{{\left( {c + a} \right)}}{b}$, $\dfrac{{\left( {a + b} \right)}}{c}$ are in AP.

Answer 1

Hint: As, it is given in the question that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in AP, it is known that whenever we add or subtract or multiply or divide any numbers which are in AP by any particular number then still those numbers will be in AP. So, first multiply these numbers with $a + b + c$, and after that subtract 1 from the numbers, then the numbers will come out still in AP.

Complete step-by-step solution:
Given, $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in AP, and we need to prove $\dfrac{{\left( {b + c} \right)}}{a}$, $\dfrac{{\left( {c + a} \right)}}{b}$, $\dfrac{{\left( {a + b} \right)}}{c}$ are in AP.
It is known that, if we multiply or divide the numbers which are in AP with a particular number, then still that number will be in AP.
Now, multiply $a + b + c$ to all the three numbers.
So, the numbers will become, $\dfrac{{a + b + c}}{a}$, $\dfrac{{a + b + c}}{b}$, $\dfrac{{a + b + c}}{c}$.
If we subtract or add some particular number from the numbers which are in AP, still these numbers will be in AP.

Now, subtract 1 from all these three numbers.
$\Rightarrow\dfrac{{a + b + c}}{a} - 1$, $\dfrac{{a + b + c}}{b} - 1$, $\dfrac{{a + b + c}}{c} - 1$, now simplify the numbers.
$\Rightarrow \dfrac{{a + b + c}}{a} - 1 = \dfrac{{a + b + c - a}}{a} = \dfrac{{b + c}}{a}$, $\dfrac{{a + b + c}}{b} - 1 = \dfrac{{a + b + c - b}}{b} = \dfrac{{a + c}}{b}$, $\dfrac{{a + b + c}}{c} - 1 = \dfrac{{a + b + c - c}}{c} = \dfrac{{a + b}}{c}$
Therefore, $\dfrac{{b + c}}{a}$, $\dfrac{{a + c}}{b}$, $\dfrac{{a + b}}{c}$ are in AP.

Note:
An arithmetic progression or arithmetic sequence the numbers are in such a sequence that the difference between consecutive numbers are the same.
Whenever we add or subtract any particular number with the numbers which are in AP, will still be in AP.
Whenever we multiply or divide any particular number with the numbers which are in AP, will still be in AP.The numbers in an AP differ by common difference among them.