First slide

Introduction to statistics

Question

Find the harmonic mean of $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \dots, \frac{n}{n + 1}$ occurring with frequencies 1, 2, 3, ..., n, respectively.

Moderate

A

$\frac{n - 1}{3 - n}$
B

$\frac{n + 1}{3 + n}$
C

$\frac{n + 1}{3 - n}$
D

None of these

Solution

We know that,
Harmonic mean $= \frac{\sum f}{\sum (\frac{f}{x})}$
$\begin{array}{l} ∴ \sum f = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ and \sum \frac{f}{x} = \frac{1}{1 / 2} + \frac{2}{2 / 3} + \frac{3}{3 / 4} + \dots + \frac{n}{n / (n + 1)} \\ = 2 + \frac{3 \times 2}{2} + \frac{4 \times 3}{3} + \dots + \frac{n (n + 1)}{n} \\ = 2 + 3 + 4 + \dots + n + (n + 1) \end{array}$
which is an arithmetic progression with a.=2 and d =1. By the formula of sum of n term of an AP,
$\begin{matrix} \sum \frac{f}{x} = [\frac{n}{2} {2 a + (n - 1) d}, we have \\ = \frac{n}{2} {2 \times 2 + n - 1} \\ = \frac{n}{2} (3 + n) \end{matrix}$
Harmonicmean= $\frac{n (n + 1) \div 2}{} n (3 + n) \div 2$ = $\frac{n + 1}{3 + n}$

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