If n harmonic means are inserted between 1 and r, then \(\frac{{{1^{st}}mean}}{{{n^{th}}mean}} = \)

Given:
Harmonic Means between 1 and r, i.e., 

Harmonic Progression (H.P.) = 1, H1, H2, H3, ... , Hn, r

Concept:
\( \frac{1}{H_n}= \frac{1}{a}+nD\)

where, \(D = \frac{\frac{1}{z}-\frac{1}{a}}{n+1}\), is the difference

here, z = last term of H.P., a = first term of H.P, n = no. of terms between a and z.

Calculation:
Here, z = r and  a = 1

For n = 1 (first mean);

\(\frac{1}{H_1}=1+\frac{1-r}{r(n+1)}\)
⇒ \(H_1=\frac{r(n+1)}{nr+1}\) ------(1)

For n = n (nth mean);
\(\frac{1}{H_n}=1+\frac{n(1-r)}{r(n+1)}\)

⇒ \(H_n=\frac{nr+r}{n+r}\)  ------(2)

from equations (1) and (2), we get

⇒\(\frac{H_1}{H_n}=\frac{\frac{nr+r}{nr+1}}{\frac{nr+r}{n+r}}\)

Therefore we get the solution,

⇒ \(\frac{H_1}{H_n}=\frac{n+r}{nr+1}\)