Question

If $H$ be the harmonic mean between $a$ and $b$, then the value of $\frac{1}{H-a}+\frac{1}{H-b}$ is

• A. $a+b$
• B. $ab$
• C. $\frac{1}{a}+\frac{1}{b}$
• D. $\frac{1}{a}-\frac{1}{b}$

Answer 1

The correct option is C

$\frac{1}{a}+\frac{1}{b}$

Explanation for the correct option:

Find the value of $\frac{1}{H-a}+\frac{1}{H-b}$:

Since, $H$ is the harmonic mean of $a$ and $b$. Then,

$H=\frac{2ab}{a-b}...\left(1\right)$

Subtract $a$ from both sides in the equation $\left(1\right)$.

$\begin{array}{rcl}H-a& =& \frac{2ab}{a+b}-a\\ & \Rightarrow & H-a=\frac{2ab-a^2-ab}{a+b}\\ & \Rightarrow & \frac{1}{H-a}=\frac{a+b}{ab-a^2}...\left(2\right)\end{array}$

Similarly, subtract $b$ from both sides in the equation $\left(1\right)$.

$\begin{array}{rcl}& \Rightarrow & H-b=\frac{2ab-b^2-ab}{a+b}\\ & \Rightarrow & \frac{1}{H-b}=\frac{a+b}{ab-b^2}...\left(3\right)\end{array}$

By adding equation $\left(2\right)$ and $\left(3\right)$, we get

$\begin{array}{rcl}\frac{1}{H-a}+\frac{1}{H-b}& =& \frac{a+b}{a\left(b-a\right)}+\frac{a+b}{b\left(a-b\right)}\\ & =& \frac{a+b}{a-b}\left[\frac{-1}{a}+\frac{1}{b}\right]\\ & =& \frac{a+b}{a-b}\left[\frac{-b+a}{ab}\right]\\ & =& \left(a+b\right)\left[\frac{1}{ab}\right]\\ & =& \frac{a}{ab}+\frac{b}{ab}\\ & =& \frac{1}{a}+\frac{1}{b}\\ & & \end{array}$

Hence, the correct option is C.