Question

If $H_1,H_2,....H_{19}$ are 19 harmonic means between 2 and 3 then $\frac{H_1+2}{H_1-2}-\frac{H_{19}+3}{H_{19}-3}=$

• A. 200
• B. 210
• C. 400
• D. 380

Answer 1

The correct option is A 200

$H_1,H_2,H_3,H_{20}$ are harmonic means between 2 and 3,

$\therefore Y_3=Y_2+(22-1)d...$ (corresponding A.P)

$\therefore Y_3-\frac{1}{2}=21d$

$\therefore -\frac{1}{6}=21d$

$\therefore d=-\frac{1}{126}$

$\therefore \frac{1}{H_1}=a+d$

$\therefore \frac{1}{H_1}=\frac{1}{2}-\frac{1}{126}$

$\therefore \frac{1}{H_1}=\frac{62}{126};H_1=\frac{126}{62}$

$\therefore H_1=\frac{63}{31}$

Similarly $H_{20}=\frac{1}{3-d}$

$\therefore \frac{H}{20}=\frac{1}{31}\frac{1}{H_{20}}=\frac{1}{3}-(-\frac{1}{126})$

$\therefore \frac{1}{H_{20}}=\frac{42+1}{126}$

$\therefore H_{20}=\frac{126}{43};\frac{H_1+2}{H_1-2}+\left(\frac{H_{20}+3}{H_{20}-3}\right)=$

$\frac{\frac{63}{31}+2}{\frac{63}{31}-2}+\frac{\frac{126}{43}+3}{\frac{126}{43}-3}=20$