Harmonic mean can be defined as a type of average. This phenomenon is majorly associated with different observations and can be defined rigidly. Harmonic mean doesn’t consider larger values and prioritises smaller values with the objective to balance the values appropriately. This article will discuss details about harmonic mean and will help students understand the formula and application of the same.

A mean is the average of the given sequence. When three non-zero quantities are in harmonic progression, the middle one is called the harmonic mean \(\left( {{\rm{HM}}} \right)\) between the other two. If \(a,\,b\) and \(c\) are in harmonic progression, then \(b\) is called the harmonic mean between \(a\) and \(c\). Harmonic mean is calculated as the average of reciprocals of a data set.

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Progression

If the terms of a sequence are written under fixed conditions, then the sequence is called progression. There are three types of progression. They are arithmetic progression, geometric progression, and harmonic progression.

Arithmetic Progression

Consider the following sequences.

(i) \(1,\,3,\,5,\,7,\,9, \ldots \ldots \)
(ii) \(4,\,8,\,16,\,24,\, \ldots \ldots \)
(iii) \(20,\,40,\,60,\,80,\, \ldots \ldots \)

In the above sequences, every term except the first is obtained by adding a fixed number (positive or negative) to the previous term. For example, in the sequence given in \((i)\), each term is acquired by adding \(2\) to the prior term. In the sequence given in \((ii)\), each term is \(4\) greater than the previous term, and in the sequence provided in \((iii)\), each term is obtained by adding \(20\) to the prior term.

All these examples are called arithmetic sequences or arithmetic progressions abbreviated as \({\rm{A}}{\rm{.P}}{\rm{.}}\)

A sequence \({a_1},\,{a_2},\,{a_3},\, \ldots \ldots \ldots ,\,{a_n}\) is called an arithmetic progression, if there is a constant term \(d\) such that

\({a_2} = {a_1} + d\)

\({a_3} = {a_2} + d\)

\({a_4} = {a_3} + d\)

……….

……….

\({a_n} = {a_{n – 1}} + d\) and so on.

There are two types of \({\rm{AP}}\) they are, finite and infinite \({\rm{AP}}\).

Finite \({\rm{AP}}\): An arithmetic progrssion having a finite number of terms is called a finite \({\rm{AP}}\). A finite \({\rm{AP}}\) has the end term.

Example: \(2,\,4,\,6,\,8,\,10\)

Infinite \({\rm{AP}}\): An \({\rm{AP}}\) which is not a finite \({\rm{AP}}\). In other words, an arithmetic progrssion which has an infinite number of terms is called an infinite \({\rm{AP}}\).

Example: \(1,\,2,\,3,\,4,\,5,\,6,\,7,\, \ldots \ldots \)

Arithmetic Mean

Given two numbers \(a\) and \(b\). We can insert a number \(A\) between them so that \(a,\,A,\,b\) is an arithmetic progression. Such a number \(A\) is called the arithmetic mean \({\rm{A}}{\rm{.M}}{\rm{.}}\) of the numbers \(a\) and \(b\). Note that, in this case, we have

\(A – a = b – A \Rightarrow A = \frac{{a + b}}{2}\)

So, we can say that the arithmetic mean between two numbers \(a\) and \(b\) as their average \(\frac{{a + b}}{2}\).

Harmonic Progressions

A sequence is said to be in harmonic progression \(\left( {{\rm{HP}}} \right)\) if the sequence formed by the reciprocals of each term are in arithemtic progression.

If the sequence \({a_1},\,{a_2},\,{a_3},\, \ldots \ldots \ldots ,\,{a_n}\) are in harmonic progression, then \(\frac{1}{{{a_1}}},\,\frac{1}{{{a_2}}},\,\frac{1}{{{a_3}}},\, \ldots \ldots \ldots ,\,\frac{1}{{{a_n}}}\) are in arithmetic progression.

Properties of Harmonic Progression

1. No term of harmonic progression can be zero.
2. The general term of harmonic progression is \({t_n} = \frac{1}{{a + (n – 1)d}}\)
3. We do not find the general formula to find out the sum of \(n\)-terms of an \({\rm{HP}}\)
4. Questions of \({\rm{HP}}\) are generally solved by reciprocating the terms and using the properties of the corresponding \({\rm{AP}}\) sequence.

Harmonic Mean

The harmonic mean \(\left( {{\rm{HM}}} \right)\) is defined as the reciprocal of the mean of the reciprocals of the statistic values. It is based on all the survey, and it is extremely defined. Harmonic mean gives less weightage to the significant values and large weightage to the small values to balance the values correctly. The harmonic Mean is used when there is a prerequisite to provide additional weight to the smaller items. It is applied in the case of times and mean rates.

When three non-zero quantities are in \({\rm{HP}}\), the middle one is called the harmonic mean \(\left( {{\rm{HM}}} \right)\) between the other two. If \(a,\,b\) and \(c\) are in \({\rm{HP}}\), then b is called the harmonic Mean between \(a\) and \(c\).

If \(a,\,b \in {R^ + }\), then harmonic mean is given by \(\frac{2}{{\frac{1}{a} + \frac{1}{b}}}\)

If \(a,\,b,\,c \in {R^ + }\), then harmonic mean is given by \(\frac{3}{{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}\)

If \({a_1},\,{a_2},\,{a_3}, \ldots \ldots \ldots ,\,{a_n} \in {R^ + }\), then harmonic mean is given by \(\frac{1}{{\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + \ldots \ldots + + \frac{1}{{{a_n}}}}}\)

Steps to Calculate the Harmonic Mean

1. Calculate the reciprocal of all the values in the given data.
2. Find the mean or average of those reciprocals by adding them and dividing them by the number of total values.
3. Then find the reciprocal of the above average which is the harmonic mean.

Insertion of \(n\) Harmonic Means Between Two Positive Real Numbers

Let two positive numbers be \(a\) and \(b\) respectively.

Let \({H_1},\,{H_2},\,{H_3}, \ldots \ldots .,\,{H_n}\) are \(n\) harmonic means inserted between two positive real numbers \(a\) and \(b\).

Then, \(a,\,{H_1},\,{H_2},\,{H_3},\, \ldots \ldots \ldots ,\,{H_n},\,b\) are in harmonic progression.

Thus, \(\frac{1}{a},\,\frac{1}{{{H_1}}},\,\frac{1}{{{H_2}}},\,\frac{1}{{{H_3}}},\, \ldots \ldots \ldots ,\,\frac{1}{{{H_n}}},\,\frac{1}{b}\) are in arithmetic progression.

Now, \(\frac{1}{b} = \frac{1}{a} + (n + 1)d\)

\( \Rightarrow \frac{1}{b} – \frac{1}{a} = (n + 1)d\)

\( \Rightarrow d = \frac{{a – b}}{{(n + 1)ab}}\)

Therefore, \(\frac{1}{{{H_1}}} = \frac{1}{a} + d = \frac{1}{a} + \frac{{a – b}}{{(n + 1)ab}}\)

Similarly, \(\frac{1}{{{H_2}}} = \frac{1}{a} + 2d = \frac{1}{a} + 2\left( {\frac{{a – b}}{{(n + 1)ab}}} \right)\)

\(\frac{1}{{{H_3}}} = \frac{1}{a} + 3d = \frac{1}{a} + 3\left( {\frac{{a – b}}{{(n + 1)ab}}} \right)\)

………….

………….

\(\frac{1}{{{H_n}}} = \frac{1}{a} + 3n = \frac{1}{a} + n\left( {\frac{{a – b}}{{(n + 1)ab}}} \right)\)

Properties of Harmonic Mean

The sum of the reciprocal of \(n\) harmonic means equals \(n\) times the single harmonic mean between the two given positive real numbers.

Let \({H_1},\,{H_2},\,{H_3},\, \ldots \ldots ..,\,{H_n}\) are \(n\) harmonic means inserted between the two given positive real numbers \(a\) and \(b\).

Since \(a,\,{H_1},\,{H_2},\,{H_3}, \ldots \ldots .,\,{H_n}\) are in harmonic progression.

 \(\frac{1}{a},\,\frac{1}{{{H_1}}},\,\frac{1}{{{H_2}}},\,\frac{1}{{{H_3}}}, \ldots \ldots \ldots ,\,\frac{1}{{{H_n}}},\,\frac{1}{b}\) are in arithmetic progression.

Therefore, \(\frac{1}{{{H_1}}} + \frac{1}{{{H_2}}} + \frac{1}{{{H_3}}} + \ldots \ldots \ldots + \frac{1}{{{H_n}}} = n \times \left( {\frac{{\frac{1}{a} + \frac{1}{b}}}{2}} \right)\)

\( = n \times \frac{1}{{\left( {\frac{2}{{\frac{1}{a} + \frac{1}{b}}}} \right)}}\)

Harmonic Mean in Statistics

Harmonic mean for ungrouped data:

Let \({x_1},\,{x_2},\,{x_3},\, \ldots \ldots \ldots ,\,{x_n}\) be the \(n\) observations then the harmonic mean is defined as

\({\rm{H}}.{\rm{M}} = \frac{n}{{\sum\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} }}\)

Harmonic mean for discrete grouped data:

\({\rm{H}}.{\rm{M}} = \frac{N}{{\sum\limits_{i = 1}^n {{f_i}} \left( {\frac{1}{{{x_i}}}} \right)}}\)

Harmonic mean for continuous data:

\({\rm{H}} \cdot {\rm{M}} = \frac{N}{{\sum\limits_{i = 1}^n {{f_i}} \left( {\frac{1}{{{x_i}}}} \right)}}\)

Where, \({{x_i}}\) is the mid-point of the class interval.

Harmonic Mean – Uses

The uses of the harmonic Mean are listed below:

1. The harmonic mean is used in finance to average multiples like the price-earnings ratio.
2. It is also used in the determination of the Fibonacci series.

Relationship Between the \({\text{AM,GM}}\) and \({\text{HM}}\)

Let \(a,\,b \in {R^ + }\)

Then, \({\rm{AM}} = (A) = \left( {\frac{{a + b}}{2}} \right)\)

\({\rm{GM}} = (G) = \sqrt {ab} \)

\({\rm{HM}} = (H) = \left( {\frac{{2ab}}{{a + b}}} \right)\)

Now, \(A \times H = \left( {\frac{{a + b}}{2}} \right) \times \left( {\frac{{2ab}}{{a + b}}} \right) = ab = {G^2}\)

\( \Rightarrow {G^2} = AH\)

\( \Rightarrow \frac{A}{G} = \frac{G}{H} \ldots \ldots (i)\)

Also, \(A – G = \frac{{a + b}}{2} – \sqrt {ab} \)

\( = \frac{1}{2}{(\sqrt a – \sqrt b )^2} \ge 0\)

\( \Rightarrow A \ge G \Rightarrow \frac{A}{G} \ge 1 \ldots \ldots (ii)\)

From \((i)\) and \((ii)\), we get

\(\frac{G}{H} = \frac{A}{G} \ge 1\)

\( \Rightarrow G \ge H\)

Hence, \(A \ge G \ge H\)

\( \Rightarrow {\rm{AM}} \ge {\rm{GM}} \ge {\rm{HM}}\)

Solved Examples – Harmonic Mean

Q.1. Find the harmonic mean for data \(2,\,4,\,6,\,8\)
Ans: Finding the reciprocal of the given values, we get
\(\frac{1}{2} = 0.5,\,\frac{1}{4} = 0.25,\,\frac{1}{6} = 0.1667,\,\frac{1}{8} = 0.125\)
The total number of statistic values is \(4\).
Therefore, average \( = \frac{{0.5 + 0.25 + 0.1667 + 0.125}}{4}\)
Average \( = 0.260425\)
Now, harmonic mean \( = \frac{1}{{{\rm{ average }}}} = \frac{1}{{0.260425}}\)
Therefore, the harmonic mean of \(2,\,4,\,6,\,8\) is \(3.84\) (approx.).

Q.2. Insert three harmonic means between \(2\) and \(3\).
Ans: Let \(a,\,b,\,c\) be the harmonic mean between \(2\) and \(3\). Then according to the relation between \({\rm{AP}}\) and \({\rm{HP}}\), we can say that \(\frac{1}{2},\,\frac{1}{a},\,\frac{1}{b},\,\frac{1}{c},\,\frac{1}{3}\) are in \({\rm{AP}}\).
We know that, \({n^{th}}\) term of an \({\rm{AP}}\) is given by \({a_n} = a + (n – 1)d\)
As we have \(\frac{1}{3}\) is the fifth term, using the above formula, we have
\(\frac{1}{3} = \frac{1}{2} + (5 – 1)d\)
\( \Rightarrow 4d = \frac{1}{3} – \frac{1}{2}\)
\( \Rightarrow 4d = \frac{{ – 1}}{6}\)
\( \Rightarrow d = \frac{{ – 1}}{{24}}\)
Now \(\frac{1}{a} = \frac{1}{2} + (2 – 1) \times \frac{{ – 1}}{{24}} \Rightarrow \frac{1}{a} = \frac{{11}}{{24}} \Rightarrow a = \frac{{24}}{{11}}\)
\(\frac{1}{b} = \frac{1}{2} + (3 – 1) \times \frac{{ – 1}}{{24}} \Rightarrow \frac{1}{b} = \frac{1}{2} – \frac{1}{{12}} \Rightarrow \frac{1}{b} = \frac{5}{{12}} \Rightarrow b = \frac{{12}}{5}\)
\(\frac{1}{c} = \frac{1}{2} + (4 – 1) \times \frac{{ – 1}}{{24}} \Rightarrow \frac{1}{c} = \frac{1}{2} – \frac{1}{8} \Rightarrow \frac{1}{c} = \frac{3}{8} \Rightarrow c = \frac{8}{3}\)
So, \(a = \frac{{24}}{{11}},\,b = \frac{{12}}{5},\,c = \frac{8}{3}\)
Therefore, the three harmonic means between \(2\) and \(3\) are \(\frac{{24}}{{11}},\,\frac{{12}}{5},\,\frac{8}{3}\)

Q.3. The following data is obtained from the survey. Compute harmonic mean.

Speed of the car	\(130\)	\(135\)	\(140\)	\(145\)	\(150\)
Number of cars	\(3\)	\(4\)	\(8\)	\(9\)	\(2\)

Ans:

\({x_i}\)	\({f_i}\)	\(\frac{{{f_i}}}{{{x_i}}}\)
\(130\)	\(3\)	\({\rm{0}}{\rm{.0231}}\)
\(135\)	\(4\)	\({\rm{0}}{\rm{.0091}}\)
\(140\)	\(8\)	\({\rm{0}}{\rm{.0571}}\)
\(145\)	\(9\)	\({\rm{0}}{\rm{.0621}}\)
\(150\)	\(2\)	\({\rm{0}}{\rm{.0133}}\)
	\(N = 26\)	Total: \({\rm{0}}{\rm{.1648}}\)

We know that \({\rm{H}}.{\rm{M}} = \frac{N}{{\sum\limits_{i = 1}^n {{f_i}} \left( {\frac{1}{{{x_i}}}} \right)}}\)
\( = \frac{{26}}{{0.1648}}\)
\({\rm{ = 157}}{\rm{.77}}\)

Therefore, the harmonic mean of the given data is \({\rm{157}}{\rm{.77}}\).

Q.4. Find the harmonic mean of the following distribution of data

Dividend Yield	\(2 – 6\)	\(6 – 10\)	\(10 – 14\)
No. of companies	\(10\)	\(12\)	\(18\)

Ans:

Class intervals	Mid-value \(\left( {{x_i}} \right)\)	No. of companies \(\left( {{f_i}} \right)\)	Reciprocal \ (\left( {\frac{1}{{{x_i}}}} \right)\)	\(\frac{{{f_i}}}{{{x_i}}}\)
\(2 – 6\)	\(4\)	\(10\)	\(\frac{1}{4}\)	\(2.5\)
\(6 – 10\)	\(8\)	\(12\)	\(\frac{1}{8}\)	\(1.5\)
\(10 – 14\)	\(12\)	\(18\)	\(\frac{1}{12}\)	\(1.5\)
		\(N = 40\)		Total \( = 5.5\)

We know that \({\rm{H}}.{\rm{M}} = \frac{N}{{\sum\limits_{i = 1}^n {{f_i}} \left( {\frac{1}{{{x_i}}}} \right)}}\)

\( = \frac{{40}}{{5.5}}\)

\({\rm{ = 7}}{\rm{.27}}\)

Therefore, the harmonic mean of the given data is \({\rm{7}}{\rm{.27}}\)

Q.5. Insert two harmonic means between \(5\) and \(11\).
Ans: Let \(a = 5\) and \(b = 11\)
Let \({H_1}\) and \({H_2}\) be the two \({\rm{HM}}\).
Therefore, \(\frac{1}{5},\,\frac{1}{{{H_1}}},\,\frac{1}{{{H_2}}},\,\frac{1}{{11}}\) are in \({\rm{AP}}\).
Let \(d\) is the common difference of this in \({\rm{AP}}\).
Therefore, \(\frac{1}{{11}} = {(n + 2)^{th}}{\rm{ term }} = {T_{n + 2}}\)
\( \Rightarrow \frac{1}{{11}} = \frac{1}{5} + [(n + 2) – 1]d\)
\( \Rightarrow d = \frac{{\frac{1}{{11}} – \frac{1}{5}}}{{(n + 1)}}\)
Therefore, for \(n = 2\)
\(d = \frac{{\frac{1}{{11}} – \frac{1}{5}}}{3} = \frac{{5 – 11}}{{55 \times 3}} = \frac{{ – 6}}{{3 \times 55}} = \frac{{ – 2}}{{55}}\)
Therefore, \({T_2} = \frac{1}{5} + (2 – 1) \times \frac{{ – 2}}{{55}} = \frac{{11 – 2}}{{55}} = \frac{9}{{55}} = \frac{1}{{{H_1}}}\)
\({T_3} = \frac{1}{5} + \left( {2 \times \frac{{ – 2}}{{55}}} \right) = \frac{7}{{55}} = \frac{1}{{{H_2}}}\)
Therefore, \({H_1} = \frac{9}{{55}}\) and \({H_2} = \frac{{55}}{7}\)

Q.6. Find the harmonic mean between \(\frac{a}{{1 – ab}}\) and \(\frac{a}{{1 + ab}}\)
Ans: Here, number of terms \(= 2\)
Let \({x_1} = \frac{a}{{1 – ab}}\) and \({x_2} = \frac{a}{{1 + ab}}\)
Then, \(\frac{1}{{{x_1}}} = \frac{{1 – ab}}{a}\) and \(\frac{1}{{{x_2}}} = \frac{{1 + ab}}{a}\)
We know that harmonic mean \( = \frac{n}{{\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}}}}\)
\( \Rightarrow \frac{2}{{\frac{{1 – ab}}{a} + \frac{{1 + ab}}{a}}}\)
\( = \frac{{2a}}{2}\)
\( = a\)
Therefore, the harmonic mean between \(\frac{a}{{1 – ab}}\) and \(\frac{a}{{1 + ab}}\) is \(a\).

Summary

The above article has learned about the harmonic mean in statistics and harmonic mean in the sequence. Also, we have learnt to insert the harmonic mean between two given numbers, uses of harmonic mean and we have learned the relationship between \({\rm{AM, GM}}\) and \({\rm{HM}}\).

Frequently Asked Questions – Harmonic Mean

Frequently asked questions related to harmonic mean is listed as follows:

Q.1. What is harmonic mean?
Ans: The Harmonic Mean \(\left( {{\rm{HM}}} \right)\) is defined as the reciprocal of the Mean of the reciprocals of the data values. In other words, when three non-zero quantities are in \({\rm{HP}}\), the middle one is called the harmonic Mean \(\left( {{\rm{HM}}} \right)\) between the other two. If \(a,\,b\) and \(c\) are in \({\rm{HP}}\), then \(b\) is called the harmonic Mean between \(a\) and \(c\).

Q.2. How harmonic mean differs from the arithmetic mean?
Ans: A sequence is said to be in Harmonic progression if the sequence formed by the reciprocals of each term are in \({\rm{AP}}\).
Let two positive numbers be \(a\) and \(b\) respectively.
Let \({H_1},\,{H_2},\,{H_3},\, \ldots \ldots ..,\,{H_n}\) are \(n\) harmonic means inserted between two positive real numbers \(a\) and \(b\).
Then, \(a,\,{H_1},\,{H_2},\,{H_3} \ldots \ldots \ldots ,\,{H_n},\,b\) are in harmonic progression.
Thus, \(\frac{1}{a},\,\frac{1}{{{H_1}}},\,\frac{1}{{{H_2}}},\,\frac{1}{{{H_3}}}, \ldots \ldots \ldots ,\,\frac{1}{{{H_n}}},\,\frac{1}{b}\) are in arithmetic progression.

Q.3. What are the advantages of harmonic mean?
Ans: The advantages of the harmonic mean are:
1. The harmonic mean is rigidly defined
2. The harmonic mean is based on all the observations of the series
3. The harmonic mean is suitable in the case of series having wide dispersion
4. The harmonic mean is suitable for further mathematical treatment
5. The harmonic mean gives less weight to large items and more weight to small items

Q.4. What is the harmonic mean of two numbers?
Ans: If \(a,\,b \in {R^ + }\), then harmonic mean of \(a\) & \(b\) is given by \(\frac{2}{{\frac{1}{a} + \frac{1}{b}}}\)

Q.5. What is geometric mean and harmonic mean?
Ans: The geometric mean of two positive numbers \(a\) and \(b\) is the number \(\sqrt {ab} \).
The harmonic mean of two positive numbers \(a\) and \(b\) is the number \(\frac{2}{{\frac{1}{a} + \frac{1}{b}}}\).

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