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

A
$n + 1$
B
$n - 1$
C
$2 n$
D
$2 n + 3$

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 $\frac{H_{n} + b}{H_{n} - b} = \frac{1 / b + 1 / H_{n}}{1 / b - 1 / H_{n}} = \frac{1 / b + 1 / H_{n}}{d}$
$∴ \frac{H_{1} + a}{H_{1} - a} + \frac{H_{n} + b}{H_{n} - b} = \frac{1 / a + 1 / H_{1}}{- d} + \frac{1 / b + 1 / H_{n}}{d}$

$= \frac{1}{d} [(\frac{1}{b} - \frac{1}{a}) + (\frac{1}{H_{n}} - \frac{1}{H_{1}})] = 2 n$

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

Solution:

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