If H1,H2,…,Hn are n harmonic means between a and b(≠a), then value of H1+aH1−a+Hn+bHn−bis equal to

# If ${H}_{1},{H}_{2},\dots ,{H}_{n}$ are $n$ harmonic means between $a$ and $b\left(\ne a\right)$, then value of $\frac{{H}_{1}+a}{{H}_{1}-a}+\frac{{H}_{n}+b}{{H}_{n}-b}$is equal to

1. A

$n+1$

2. B

$n-1$

3. C

$2n$

4. D

$2n+3$

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### Solution:

As are in H.P.
$\frac{1}{a},\frac{1}{{H}_{1}},\frac{1}{{H}_{2}}\cdots \frac{1}{{H}_{n}},\frac{1}{b}$ are in A.P.
Let $d$be the common difference of this A.P., then
$\frac{1}{b}=\frac{1}{a}+\left(n+1\right)d$
and  $\frac{1}{{H}_{n}}-\frac{1}{{H}_{1}}=\left(n-1\right)d$
Now, $\frac{{H}_{1}+a}{{H}_{1}-a}=\frac{1/a+1/{H}_{1}}{1/a-1/{H}_{1}}=\frac{1/a+1/{H}_{1}}{-d}$

and $\frac{{H}_{n}+b}{{H}_{n}-b}=\frac{1/b+1/{H}_{n}}{1/b-1/{H}_{n}}=\frac{1/b+1/{H}_{n}}{d}$

$=\frac{1}{d}\left[\left(\frac{1}{b}-\frac{1}{a}\right)+\left(\frac{1}{{H}_{n}}-\frac{1}{{H}_{1}}\right)\right]=2n$