Question

If $H_1,H_2,...,H_n$ are the $n$ harmonic means between $a$ and $b$, then $\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=$

• A. $n$
• B. $n+1$
• C. $2n$
• D. $2n-2$

Answer 1

Answer: (C)

$2n$

Given $H_1,H_2,...,H_n$ are the $n$ harmonic means between $a$ and $b$,

$\therefore {\frac{1}{H}}_n=\frac{1}{b}-d,\frac{1}{H_1}=\frac{1}{a}+d$

$\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=\frac{\frac{1}{a}+\frac{1}{H_1}}{\frac{1}{a}-\frac{1}{H_1}}+\frac{\frac{1}{b}+\frac{1}{H_n}}{\frac{1}{b}-\frac{1}{H_n}}$

$=\frac{\frac{1}{a}+\frac{1}{a}+d}{\frac{1}{a}-d-\frac{1}{a}}+\frac{\frac{1}{b}+\frac{1}{b}-d}{\frac{1}{b}+d-\frac{1}{b}}=\frac{\frac{2}{a}+d}{-d}+\frac{\frac{2}{b}-d}{d}$

$=\frac{\frac{2}{b}-\frac{2}{a}}{d}-2=2(n+1)-2[\because \frac{1}{b}=\frac{1}{a}+(n+2-1)d]=2n$