If $H_1,H_2,\ldots,H_n$ are $n$ harmonic means between $a$ and $b$, $a\ne b$, then the value of $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b}$ is equal to

Step 1: Convert HP to AP
Since $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.

Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$

Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$

$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$

Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$

Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $