Question

If the terms $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ are in A.P. Prove that a, b, c are in A.P.

Answer 1

Hint: We first try to find if the conditions for the terms $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ satisfy. For $x,y,z$ are in A.P, we can say that $z-y=y-x$. We also can prove that part with the use of binary operations for $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$.

Complete step-by-step solution:
The terms $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ are in A.P.
If the terms $x,y,z$ are in A.P. then we can say that $z-y=y-x$.
Therefore, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)-b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)=b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)-a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$.
We now simplify the equation to get
$\begin{align}
  & c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)-b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)=b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)-a\left( \dfrac{1}{b}+\dfrac{1}{c} \right) \\
 & \Rightarrow \dfrac{c}{a}+\dfrac{c}{b}-\dfrac{b}{c}-\dfrac{b}{a}=\dfrac{b}{c}+\dfrac{b}{a}-\dfrac{a}{b}-\dfrac{a}{c} \\
\end{align}$
Now based on the denominator value we take the common terms out.
$\begin{align}
  & \dfrac{c}{a}+\dfrac{c}{b}-\dfrac{b}{c}-\dfrac{b}{a}=\dfrac{b}{c}+\dfrac{b}{a}-\dfrac{a}{b}-\dfrac{a}{c} \\
 & \Rightarrow \left( \dfrac{c}{a}-\dfrac{b}{a} \right)+\left( \dfrac{c}{b}-\dfrac{b}{c} \right)=\left( \dfrac{b}{c}-\dfrac{a}{c} \right)+\left( \dfrac{b}{a}-\dfrac{a}{b} \right) \\
\end{align}$
We complete the simplification of the dfractions in the brackets and get
$\begin{align}
  & \left( \dfrac{c}{a}-\dfrac{b}{a} \right)+\left( \dfrac{c}{b}-\dfrac{b}{c} \right)=\left( \dfrac{b}{c}-\dfrac{a}{c} \right)+\left( \dfrac{b}{a}-\dfrac{a}{b} \right) \\
 & \Rightarrow \left( \dfrac{c-b}{a} \right)+\left( \dfrac{{{c}^{2}}-{{b}^{2}}}{bc} \right)=\left( \dfrac{b-a}{c} \right)+\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{ab} \right) \\
\end{align}$
We now can take common $\left( c-b \right)$ from left side and $\left( b-a \right)$ from right side
$\begin{align}
  & \left( \dfrac{c-b}{a} \right)+\left( \dfrac{{{c}^{2}}-{{b}^{2}}}{bc} \right)=\left( \dfrac{b-a}{c} \right)+\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{ab} \right) \\
 & \Rightarrow \left( c-b \right)\left[ \dfrac{1}{a}+\dfrac{c+b}{bc} \right]=\left( b-a \right)\left[ \dfrac{1}{c}+\dfrac{b+a}{ab} \right] \\
\end{align}$
We have one common term on both sides as the simplification as we get
\[\left[ \dfrac{1}{a}+\dfrac{c+b}{bc} \right]=\dfrac{bc+ac+ab}{abc}=\left[ \dfrac{1}{c}+\dfrac{b+a}{ab} \right]\]
Therefore, we can omit that from both sides and get $\left( c-b \right)=\left( b-a \right)$ which gives a, b, c are in A.P.

Note: We can also use the simple conditions of A.P. to prove that a, b, c are in A.P.
We know that any binary operation of same number with the A.P. numbers won’t change the conditions of the A.P.
For our given terms $a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)$ are in A.P.
We add 1 to all of the them and they still remain in A.P.
$a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)+1=a\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)$, $b\left( \dfrac{1}{c}+\dfrac{1}{a} \right)+1=b\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)$, $c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)+1=c\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)$ are in A.P.
We now divide with $\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)$ and still they remain as A.P.
Therefore, a, b, c are in A.P.