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# Harmonic Mean for IIT JEE

A Harmonic Progression is a sequence if the reciprocals of its terms are in Arithmetic Progression, and harmonic mean (or shortly written as HM) can be calculated by dividing the number of terms by reciprocals of its terms.

For example,

$$\begin{array}{l}\text{Terms}\ t_1,\ t_2,\ t_3\ \text{are in HP if and only if}\ \frac{1}{{{t}_{1}}},\frac{1}{{{t}_{2}}},\frac{1}{{{t}_{3}}},…\ \text{are in AP.}\end{array}$$

Then,

$$\begin{array}{l}\text{Harmonic Mean} = \frac{3}{\frac{1}{t_1} + \frac{1}{t_2}+\frac{1}{t_3}}\end{array}$$

HM gives less weightage to large values and more weightage to small values and thus does the balancing act properly. The Harmonic mean has an application in many fields like physics, finance, geometry, trigonometry etc. Harmonic Mean is used when we need to give greater weights to smaller items.

Table of Content:

## What is Harmonic Mean?

The harmonic mean in statistics is the reciprocal of the arithmetic mean of the values. It is based on all observations and is rigidly defined. It is applied in the case of times and average rates.

### Relationship between Arithmetic mean, Geometric Mean and Harmonic Mean

For n terms,  a1, a2, a3, ……., an.

Arithmetic mean

$$\begin{array}{l}=\frac{a_{1}+a_{2}+a_{3}+….+a_{n}}{n}\end{array}$$

Geometric mean

$$\begin{array}{l}=\sqrt[n]{a_{1}.a_{2}.a_{3}….a_{n}}\end{array}$$

Harmonic mean

$$\begin{array}{l}=\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+…+\frac{1}{a_{n}}}\end{array}$$

In statistics, Arithmetic Mean, Geometric Mean and Harmonic Mean are called Pythagorean Means.

 Relationship: $$\begin{array}{l}G = \sqrt{A\times H}\end{array}$$

Where Arithmetic mean is denoted as A, Geometric Mean as G and Harmonic Mean as H.

Related Articles:

## Harmonic Mean Formula

 Formula: If x1,x2,….,xn are the n individual items, the Harmonic mean is given by Harmonic Mean $$\begin{array}{l}=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+….\frac{1}{x_n}}\end{array}$$

## Harmonic Mean of Two Numbers

Harmonic Mean of two numbers is an average of two numbers.

In particular, Let a and b be two given numbers and H be the HM between them a, H, b are in HP.

Hence,

$$\begin{array}{l}H=\frac{2}{\frac{1}{a}+\frac{1}{b}}\,\,\,i.e.,\,\,\,H=\frac{2ab}{(a+b)}\end{array}$$

Again, if three terms are in HP, then the middle term is called the Harmonic Mean between the other two, so if a, b, c are in HP, then b is the HM of a and c.

Insert n-Harmonic Mean Between Two numbers

Let a and b be two given numbers and H1, H2, H3,….., Hn are n HM’s between them.

Then, a, H1, H2, H3, …., Hn, b will be in HP, if D be the common difference of the corresponding AP.

b = (n+2)th term of HP.

That is,

$$\begin{array}{l}\frac{1}{(n+2)th\,term\,of\,corresponding\,AP}\end{array}$$
$$\begin{array}{l}=\frac{1}{\frac{1}{a}+(n+2-1)D}\end{array}$$

This implies, D = (1/b – 1/a)/(n+1)

Therefore,

$$\begin{array}{l}\frac{1}{H_1} = \frac{1}{a} + D, \frac{1}{H_2} = \frac{1}{a} + 2D, …., \frac{1}{H_n} = \frac{1}{a} + nD\end{array}$$
,

$$\begin{array}{l}\frac{1}{{{H}_{1}}}=\frac{1}{a}+D,\frac{1}{{{H}_{2}}}=\frac{1}{a}+2D,…,\frac{1}{{{H}_{n}}}=\frac{1}{a}+nD\end{array}$$
,

$$\begin{array}{l}\frac{1}{{{H}_{1}}}=\frac{1}{a}+\frac{(a-b)}{ab(n+1)},\frac{1}{{{H}_{2}}}=\frac{1}{a}+\frac{2(a-b)}{ab(n+1)},..,\frac{1}{{{H}_{n}}}=\frac{1}{a}+\frac{n(a-b)}{ab(n+1)}\end{array}$$
,

## Single HM of n Positive Numbers:

Let n positive numbers be a1, a2, a3, …, an and H be the HM of these numbers, then

$$\begin{array}{l}H=\frac{n}{\left(\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+…+\frac{1}{{{a}_{n}}} \right)}\end{array}$$

Points to Remember :

1. HM of a, b, c is

$$\begin{array}{l}\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\,\,or\,\,\frac{3abc}{ab+bc+ca}\end{array}$$

2. The AM between two numbers a and b is (a + b)/2.

It does not follow that HM between the same numbers is 2/(a + b).

$$\begin{array}{l}\text{The HM is the reciprocals of}\ \frac{\frac{1}{a}+\frac{1}{b}}{2},\ \text{i.e.,}\ \frac{2ab}{(a+b)}.\end{array}$$

## Harmonic Mean For Grouped Data

Consider x1, x2, x3, …. ,xn are n individual values and f1, f2, f3, …..,fn are the frequencies, then

$$\begin{array}{l}HM = \frac{f_{1}+f_{2}+f_{3}+…+f_{n}}{\frac{f_{1}}{x_{1}}+\frac{f_{2}}{x_{2}}+\frac{f_{3}}{x_{3}}+…+\frac{f_{n}}{x_{n}}}\end{array}$$

$$\begin{array}{l}=\frac{\sum f}{\sum (\frac{f}x{})}\end{array}$$

## Weighted Harmonic Mean

The Harmonic Mean as defined is the special case,

When all of the weights are equal to 1, and

Equivalent to any weighted HM considering all weights are equal.

 Definition: If a set of weights w1, w2,……………,wn is associated with the sample space x1, x2,……….…, xn, the Weighted Harmonic Mean is defined by $$\begin{array}{l}HM_w = \frac{\sum_{i=1}^{n}w_i}{\sum_{i=1}^{n}\frac{w_i}{x_i}}\end{array}$$

### How to Calculate the Harmonic Mean

Below are Steps to find the harmonic mean of any data:

Step 1: Understand the given data and arrange it.

Step 2: Set up the harmonic mean formula (Given above)

Step 3: Plug the value of n and sum of reciprocal of all the entries into the formula.

Step 4: Solve and get your result.

## Harmonic Mean Problems

Example 1: Let H be the harmonic mean between x and y iff

$$\begin{array}{l}\frac{H+x}{H-x}+\frac{H+y}{H-y}=2\end{array}$$

Solution: H be the harmonic mean between x and y, then

$$\begin{array}{l}H=\frac{2xy}{x+y}\end{array}$$
$$\begin{array}{l}\frac{H}{x}=\frac{2y}{x+y}\ and\ \frac{H}{y}=\frac{2x}{x+y}\end{array}$$

By componendo and dividendo, we have

$$\begin{array}{l}\frac{H+x}{H-x}=\frac{2y+x+y}{2y-x-y}=\frac{x+3y}{y-x}\end{array}$$
and
$$\begin{array}{l}\frac{H+y}{H-y}=\frac{2x+x+y}{2x-x-y}=\frac{3x+y}{x-y}\end{array}$$
$$\begin{array}{l}\frac{H+x}{H-x}+\frac{H+y}{H-y}=\frac{x+3y}{y-x}+\frac{3x+y}{x-y}=\frac{x+3y-3x-y}{y-x}=\frac{2(y-x)}{(y-x)}=2\end{array}$$

Now,

$$\begin{array}{l}\frac{H+x}{H-x}+\frac{H+y}{H-y}=2\end{array}$$
$$\begin{array}{l}\Rightarrow \left( \frac{H+x}{H-x}-1 \right)=\left( 1-\frac{H+y}{H-y} \right)\Rightarrow \frac{2x}{H-x}=\frac{-2y}{H-y}\end{array}$$

i.e., Hx – xy = -Hy + xy

H(x + y) = 2xy

$$\begin{array}{l}\Rightarrow H=\frac{2xy}{(x+y)}\end{array}$$

This is true as x, H, and y are in HP. Hence, the required result.

Example 2: Find the harmonic mean for integers from 15 to 24.

Solution: There are 10 integers between 15 to 24 i.e. n = 10

Let x1 = 15, x2 = 16, …………..…, x10 = 24

Sum of reciprocal of all the terms = (1/15) + (1/16) + (1/17) + (1/18) + (1/19) + (1/20) + (1/21) + (1/22) + (1/23) + (1/24) + (1/25) = 1/1.906

Therefore the harmonic mean is:

HM = (number of terms) / (Sum of reciprocal of all the terms)

= 10/(1/1.906)

= 19.06

Example 3: If H is the harmonic mean between p and q, then the value of (H / p) + (H / q) is

$$\begin{array}{l}H=\frac{2 p q}{p+q} \\ \therefore \frac{H}{p}+\frac{H}{q}\\=\frac{2 q}{p+q}+\frac{2 p}{p+q}=\frac{2(p+q)}{p+q}\\=2\end{array}$$

Example 4: If the harmonic mean between a and b be H, then the value of [1 /(H − a)] + [1 /(H − b)] is

$$\begin{array}{l}\begin{array}{l} \text { Putting } H=\frac{2 a b}{a+b} \\ \text { We have } \frac{1}{H-a}+\frac{1}{H-b} \\=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}\\=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}\\ =\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)\\=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)\\=\frac{a+b}{a b}\\=\frac{1}{a}+\frac{1}{b} \end{array}\end{array}$$

Example 5:

$$\begin{array}{l}\text{If}\ \frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\ \text{is the harmonic mean between a and b,}\end{array}$$
then the value of n is

$$\begin{array}{l}\begin{array}{l} \text { We have } \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\\=\frac{2 a b}{a+b}\\ \Rightarrow a^{n+2}+a b^{n+1}+b a^{n+1}+b^{n+2}\\=2 a^{n+1} b+2 b^{n+1} a \\ \Rightarrow a^{n+1}(a-b)\\=b^{n+1}(a-b)\\ \text { or } \\ \left(\frac{a}{b}\right)^{n+1}=(1)=\left(\frac{a}{b}\right)^{0} \end{array}\end{array}$$

Hence, n = -1

Example 6: If the harmonic mean between a and b be H, then (H + a) / (H − a) + (H + b) / (H − b) =

$$\begin{array}{l}\text{Put}\ H=\frac{2 a b}{a+b}\ \text{in}\ \frac{H+a}{H-a}+\frac{H+b}{H-b}\\=\frac{2\left(H^{2}-a b\right)}{(H-a)(H-b)}\\=\frac{2\left[\frac{4 a b}{(a+b)^{2}}-a b\right]}{\left[\frac{4 a b}{(a+b)^{2}}-a b\right]}\\ =2 \\ Trick: \text{Let}\ a=1, \ H=\frac{1}{2} \ and \ b=\frac{1}{3} \ then \ \frac{H+a}{H-a}+\frac{H+b}{H-b}\\=\frac{3 / 2}{-1 / 2}+\frac{5 / 6}{1 / 6}\\=2\end{array}$$

Example 7: The harmonic mean of a / (1 − ab) and a / (1 + ab) is

$$\begin{array}{l}H . M=\frac{2\left(\frac{a^{2}}{1-a^{2} b^{2}}\right)}{\frac{a}{1-a b}+\frac{a}{1+a b}}\\=\frac{2 a^{2}}{2 a}\\=a\end{array}$$

## Frequently Asked Questions on Harmonic Mean

### What do you mean by Harmonic Mean?

Harmonic Mean is the reciprocal of the arithmetic mean of the values. It is calculated by dividing the number of observations by the reciprocal of each number in the series.

### Give the relation between Arithmetic Mean, Geometric Mean and Harmonic Mean.

Arithmetic Mean (A), Geometric Mean(G) and Harmonic Mean (H) are related by the formula, G2 = AH.

### Give the formula for the Harmonic Mean of two numbers a and b.

Harmonic Mean of a and b is given by HM = 2ab/(a+b).

### How to calculate the Harmonic Mean?

We have to find the sum of the reciprocals of each term in the given data set. Then divide the total number of terms in the data set by this value.

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