If n harmonic means between a and b is given as ${{H}_{1}},{{H}_{2}},....{{H}_{n}}$ , then value of $\dfrac{{{H}_{1}}+a}{{{H}_{1}}-a}+\dfrac{{{H}_{n}}+b}{{{H}_{n}}-b}$ is
(A). n
(B). $n-1$
(C). 2n
(D). $2n-2$

Hint: First, here we will convert ${{H}_{1}},{{H}_{2}},....{{H}_{n}}$ these values in reciprocal form and will considered it as Arithmetic progression (A.P.) . Then we will use formula for finding ${{n}^{th}}$ formula i.e. ${{T}_{n}}=a+\left( n-1 \right)d$ . From this we will get common difference value d. And then we will find the value of ${{H}_{1}}$ and ${{H}_{n}}$ . Then substituting all the values in the given equation in question, thus required answers will be obtained.

Complete step-by-step solution -
Harmonic Mean (H.P): The harmonic mean is the reciprocal of the average of the reciprocals.
The formula is given by Harmonic mean $=\dfrac{n}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+.....}$
Where a, b, c are the values and n is how many values.
So, here we have ${{H}_{1}},{{H}_{2}},....{{H}_{n}}$ in harmonic progression which is total n values.
So, as per the formula we can write reciprocal of n values as:
$\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{{{H}_{3}}},.....\dfrac{1}{{{H}_{n}}}$ ……………………..(1)
Now, we can see that all are in sequence which can be considered as Arithmetic progression (A.P.)
So, equation (1) is in A.P. between a and b values. So, we can rewrite it as
$a,{{H}_{1}},{{H}_{2}},....{{H}_{n}},b$
So, again using the H.P. formula we get it as
 $\dfrac{1}{a},\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{{{H}_{3}}},.....\dfrac{1}{{{H}_{n}}},\dfrac{1}{b}$ ……………………..(2)
Here a and b are added to the n terms. So, the value of total terms will be n+2.
Now, we will apply ${{n}^{th}}$ formula of A.P. which is given as ${{T}_{n}}=a+\left( n-1 \right)d$ where a is the first term i.e. $\dfrac{1}{a}$ , d is the common difference between two consecutive terms and ${{T}_{n}}$ is last term in series i.e. $\dfrac{1}{b}$ and n equals to n+2. So, applying this formula, we get
${{T}_{n}}=a+\left( n-1 \right)d$
$\dfrac{1}{b}=\dfrac{1}{a}+\left( n-1+2 \right)d$
$\dfrac{1}{b}=\dfrac{1}{a}+\left( n+1 \right)d$
So, on solving and takin LCM we get value of d as:
$d=\dfrac{a-b}{ab\left( n+1 \right)}$ ……………………….(3)
Now, we will find the value of $\dfrac{1}{{{H}_{1}}}$ which will be equal to the first term plus common difference. So, we will get $\dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+d$ . Putting the values in this equation, we will get
$\dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+d\Rightarrow \dfrac{1}{a}+\dfrac{a-b}{ab\left( n+1 \right)}$
On taking LCM, we will get
$\dfrac{1}{{{H}_{1}}}=\dfrac{b\left( n+1 \right)+a-b}{ab\left( n+1 \right)}$
$\dfrac{1}{{{H}_{1}}}=\dfrac{bn+b+a-b}{ab\left( n+1 \right)}=\dfrac{bn+a}{ab\left( n+1 \right)}$
$\therefore {{H}_{1}}=\dfrac{ab\left( n+1 \right)}{bn+a}$ ……………………..(4)
Similarly, we will find the value of $\dfrac{1}{{{H}_{n}}}=\dfrac{1}{a}+nd$ . So, on solving we will get the answer as
$\dfrac{1}{{{H}_{n}}}=\dfrac{an+b}{ab\left( n+1 \right)}$
 $\therefore {{H}_{n}}=\dfrac{ab\left( n+1 \right)}{an+b}$ ………………………………..(5)
Now, substituting all the values in given equation, we get
$\dfrac{{{H}_{1}}+a}{{{H}_{1}}-a}+\dfrac{{{H}_{n}}+b}{{{H}_{n}}-b}$
$\Rightarrow \dfrac{\dfrac{ab\left( n+1 \right)}{bn+a}+a}{\dfrac{ab\left( n+1 \right)}{bn+a}-a}+\dfrac{\dfrac{ab\left( n+1 \right)}{an+b}+b}{\dfrac{ab\left( n+1 \right)}{an+b}-b}$
Now taking LCM and cancelling the denominator term, we get
$\Rightarrow \dfrac{\left( 2n+1 \right)ab+{{a}^{2}}}{ab-{{a}^{2}}}+\dfrac{\left( 2n+1 \right)ab+{{b}^{2}}}{ab-{{b}^{2}}}$
Now takin a common from first term and b common from second term and cancelling out, we get as
$\Rightarrow \dfrac{\left( 2n+1 \right)b+a}{b-a}+\dfrac{\left( 2n+1 \right)a+b}{a-b}$
Taking LCM of the denominator and minus sign common from second term, we get
$\Rightarrow \dfrac{\left( 2n+1 \right)b+a-\left( 2n+1 \right)a-b}{b-a}$
On further simplification and opening the brackets, we get
$\Rightarrow \dfrac{2nb-2na}{b-a}=\dfrac{\left( b-a \right)2n}{b-a}=2n$
So, the value of $\dfrac{{{H}_{1}}+a}{{{H}_{1}}-a}+\dfrac{{{H}_{n}}+b}{{{H}_{n}}-b}$ is 2n.
Hence, option (c) is correct.

Note: Don’t forget to consider harmonic progression values as arithmetic progression and then applying ${{n}^{th}}$ term formula which is given as ${{T}_{n}}=a+\left( n-1 \right)d$ . Also, remember to add 2 values in n terms of values which is value a and b otherwise the answer will not be correct and on solving all the steps will be in H terms i.e. harmonic terms and will get so much complex. So, be careful with this type of questions.