Question

If b is the harmonic mean between a and c, prove that $\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}$.

Answer 1

Hint: We first use the condition of HP for the given numbers where $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$. We simplify the condition and multiply with $abc$. Then we form the equation taking difference of two numbers to find the given form of $\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}$.

Complete step by step solution:
As b is the harmonic mean between a and c, we can write $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$.
We simplify the expression by multiplying with $abc$ and get
$\begin{align}
  & abc\left( \dfrac{1}{a}+\dfrac{1}{c} \right)=\dfrac{2abc}{b} \\
 & \Rightarrow bc+ab=2ac \\
\end{align}$
We can write the equation as
$\begin{align}
  & bc+ab=2ac \\
 & \Rightarrow ac-bc=ab-ac \\
\end{align}$
We take the common terms and get
$\begin{align}
  & ac-bc=ab-ac \\
 & \Rightarrow -c\left( b-a \right)=a\left( b-c \right) \\
\end{align}$
We now write them as dfraction form and get
$\begin{align}
  & -c\left( b-a \right)=a\left( b-c \right) \\
 & \Rightarrow \dfrac{a}{b-a}=\dfrac{-c}{b-c} \\
\end{align}$
We multiply $\dfrac{1}{b}$ both sides to get
$\begin{align}
  & \dfrac{a}{b-a}=\dfrac{-c}{b-c} \\
 & \Rightarrow \dfrac{a}{b\left( b-a \right)}=\dfrac{-c}{b\left( b-c \right)} \\
\end{align}$
We form the numerators with respect to the denominators and get
$\begin{align}
  & \dfrac{a}{b\left( b-a \right)}=\dfrac{-c}{b\left( b-c \right)} \\
 & \Rightarrow \dfrac{b-\left( b-a \right)}{b\left( b-a \right)}=\dfrac{\left( b-c \right)-b}{b\left( b-c \right)} \\
\end{align}$
Simplifying we get
$\begin{align}
  & \dfrac{b-\left( b-a \right)}{b\left( b-a \right)}=\dfrac{\left( b-c \right)-b}{b\left( b-c \right)} \\
 & \Rightarrow \dfrac{1}{\left( b-a \right)}-\dfrac{1}{b}=\dfrac{1}{b}-\dfrac{1}{\left( b-c \right)} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{2}{b} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} \\
\end{align}$
Thus proved the given expression.

Note: A sequence of numbers is said to be a harmonic progression if the reciprocal of those numbers is in AP. We actually convert the given numbers into their AP form to find the given expression of $\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}$ as $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP. The easier way to solve the problem is to go for back process and get the HP relation of the given numbers from $\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}$ and then solve the problem the way it is done in the solution part.