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

As a,H1,H2,…,Hnb are in H.P. 1a,1H1,1H2⋯1Hn,1b are in A.P.Let dbe the common difference of this A.P., then 1b=1a+(n+1)dand 1Hn−1H1=(n−1)dNow, H1+aH1−a=1/a+1/H11/a−1/H1=1/a+1/H1−dand Hn+bHn−b=1/b+1/Hn1/b−1/Hn=1/b+1/Hnd∴ H1+aH1−a+Hn+bHn−b=1/a+1/H1−d+1/b+1/Hnd =1d1b−1a+1Hn−1H1=2n

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

As a,H1,H2,…,Hnb are in H.P. 1a,1H1,1H2⋯1Hn,1b are in A.P.Let dbe the common difference of this A.P., then 1b=1a+(n+1)dand 1Hn−1H1=(n−1)dNow, H1+aH1−a=1/a+1/H11/a−1/H1=1/a+1/H1−dand Hn+bHn−b=1/b+1/Hn1/b−1/Hn=1/b+1/Hnd∴ H1+aH1−a+Hn+bHn−b=1/a+1/H1−d+1/b+1/Hnd =1d1b−1a+1Hn−1H1=2n

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If $H_{1}, H_{2}, \dots, H_{n}$ are $n$ harmonic means between $a$ and $b (\neq a)$ , then value of $\frac{H_{1} + a}{H_{1} - a} + \frac{H_{n} + b}{H_{n} - b}$ is equal to

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$n + 1$

$n - 1$

$2 n$

$2 n + 3$

answer is C.

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Detailed Solution

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