Question

If$H_1,H_2,....H_{20}$be 20 harmonic means between 2 and 3, then$\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=$

• A. 20
• B. 21
• C. 40
• D. 38

Answer 1

The correct option is C

40

$\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=\frac{\frac{1}{2}+\frac{1}{H_1}}{\frac{1}{2}-\frac{1}{H_1}}+\frac{\frac{1}{3}+\frac{1}{H_{20}}}{\frac{1}{3}-\frac{1}{H_{20}}}$

$=\frac{\frac{1}{2}+\frac{1}{2}+d}{\frac{1}{2}-d-\frac{1}{2}}+\frac{\frac{1}{3}+\frac{1}{3}-d}{\frac{1}{3}+d-\frac{1}{3}}$

$\frac{\frac{2}{2}+d}{-d}+\frac{\frac{2}{3}-d}{d}$

$\frac{\frac{2}{3}-1}{d}-2$

$=2\times 21-2[asalso,\frac{1}{3}=\frac{1}{2}+21d]$
$=40$