Question

Let 'n' Arithmetic Means and 'n' Harmonic Means are inserted between two positive number 'a' and 'b'. If sum of all Arithmetic Means is equal to sum of reciprocal all Harmonic means, then product of number is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. 3

Answer 1

The correct option is A 1

$a_1,A_1,A_2.......A_n,b$

$A_1,A_2.......A_n,$ are arithmetic means between a and b.

$b=a+(n+1)d$

$\begin{array}{c}d=\frac{b-a}{n+1}\\ \frac{n}{2}\left[A_1+A_n\right]=\frac{n}{2}\left[2A_1+\left(n-1\right)d\right]\\ =\frac{n}{2}\left[2\left(a+d\right)+\left(n-1\right)d\right]\\ =\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ =\frac{n}{2}\left[2a+b-a\right]\\ =\frac{n}{2}\left[b+a\right]\\ a_1,H_1,H_2......H_{n}bareinH.P\end{array}$

$H_1,H_2......H_n$ are harmonic means between $a$ & $b$

$\begin{array}{c}\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2}.........\frac{1}{H_n}areinA.P\\ \frac{1}{b}=\frac{1}{a}+\left(n+1\right)d_2\\ \frac{1}{b}-\frac{1}{a}=\left(n+1\right)d_2\\ \frac{a-b}{ab}=\left(n+1\right)d_2\end{array}$

$\begin{array}{c}=\frac{1}{H_1},\frac{1}{H_2}.........\frac{1}{H_n}\\ =\frac{n}{2}\left[\frac{1}{H_2}+\frac{1}{H_n}\right]\\ =\frac{n}{2}\left[\frac{2}{H_1}+\left(n-1\right)d_2\right]\\ =\frac{n}{2}\left[\frac{2}{a}+\left(n-1\right)d_2\right]\\ =\frac{n}{2}\left[\frac{2}{a}+\frac{a-b}{ab}\right]\\ =\frac{n}{2}\frac{a+b}{ab}\\ \frac{n}{2}\frac{a+b}{ab}=\frac{n}{2}\left(a+b\right)\\ ab=1\end{array}$

Option $A$ is correct answer.