Question

If $H_1,H_2,\dots ,H_{20}$ be $20$ harmonic means between $2$ and $3$, then the value of $\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}$ is

• A. $20$
• B. $21$
• C. $40$
• D. $41$

Answer 1

The correct option is C $40$

We know that,
$\frac{1}{3}=\frac{1}{2}+21d\Rightarrow d=-\frac{1}{6\times 21}\cdots \left(1\right)$
Now, given
$\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=\frac{\frac{1}{2}+\frac{1}{H_1}}{\frac{1}{2}-\frac{1}{H_1}}+\frac{\frac{1}{3}+\frac{1}{H_{20}}}{\frac{1}{3}-\frac{1}{H_{20}}}$
$=\frac{\frac{1}{2}+\frac{1}{2}+d}{\frac{1}{2}-d-\frac{1}{2}}+\frac{\frac{1}{3}+\frac{1}{3}-d}{\frac{1}{3}-\frac{1}{3}+d}(\because \frac{1}{H_1}=\frac{1}{2}+d,\frac{1}{H_{20}}=\frac{1}{3}-d)$
$=\frac{1+d}{-d}+\frac{\frac{2}{3}-d}{d}=\frac{\frac{2}{3}-d-1-d}{d}=\frac{-\frac{1}{3}-2d}{d}=-\frac{1}{3d}-2$
$=\frac{6\times 21}{3}-2=40$ $[$Using equation $\left(1\right)$$]$