Question

If $H_1,H_2,...,H_{20}$ 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 $40$

Given,

The sequence is in harmonic progression between $2$ and $3$ and $n=20$

Now,

$d=\frac{\frac{1}{b}-\frac{1}{a}}{n+1}=\frac{\frac{1}{3}-\frac{1}{3}}{20+1}=\frac{-1}{126}$

$\frac{1}{H_1}=\frac{1}{a}+d=\frac{1}{2}-\frac{1}{126}=\frac{62}{126}$

$H_1=\frac{126}{62}$

Similarly,

$H_{2}O=\frac{1}{a}+20d$

$\therefore \frac{H_1+2}{H_2-2}+\frac{H_{2}O+3}{H_{2}O-3}=\frac{\frac{126}{62}+2}{\frac{126}{62}-2}+\frac{\frac{126}{43}+3}{d\frac{126}{43}-3}$

$=125+(-85)=40$

Hence,

Option $C$ is correct answer.