If $H_1,H_2,H_3\cdots H_n$ be $n$ harmonic means between $a$ and $b$ then $\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=$

$\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{H_3}\cdots ,\frac{1}{H_n},\frac{1}{b}$  in A.P with $n+2$ terms

$t_n=a+\left(n-1\right)d$

$\frac{1}{b}=\frac{1}{a}+\left(n+2-1\right)d\Rightarrow d=\frac{a-b}{ab\left(n+1\right)}$

$\frac{1}{H_1}=\frac{1}{a}+d=\frac{1}{a}+\frac{a-b}{ab\left(n+1\right)}$

$\therefore H_1=\frac{ab\left(n+1\right)}{a+bn}$

$\frac{1}{H_n}=\frac{1}{a}+nd=\frac{1}{a}+n\frac{\left(a-b\right)}{ab\left(n+1\right)}$ $=\frac{an+b}{ab\left(n+1\right)}$

$H_n=\frac{ab\left(n+1\right)}{b+an}$

$\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=\frac{\frac{ab\left(n+1\right)}{a+bn}+a}{\frac{ab\left(n+1\right)}{a+bn}-a}+\frac{\frac{ab\left(n+1\right)}{b+an}+b}{\frac{ab\left(n+1\right)}{b+an}-b}$

$=\frac{abn+ab+a^2+abn}{abn+ab-a^2-abn}+\frac{abn+ab+b^2+abn}{abn+ab-b^2-abn}$

$=\frac{2abn+ab+a^2}{ab-a^2}+\frac{2abn+ab+b^2}{ab-b^2}$

$=\frac{2bn+b+a}{\left(b-a\right)}+\frac{2an+a+b}{a-b}$

$=\frac{-2bn-b-a+2an+a+b}{\left(a-b\right)}$

$=\frac{2n\left(a-b\right)}{\left(a-b\right)}=2n$