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, Terms t1, t2, t3 is HP if and only if 1t1,1t2,1t3,\frac{1}{{{t}_{1}}},\frac{1}{{{t}_{2}}},\frac{1}{{{t}_{3}}},… is an AP.

Then, Harmonic Mean = 31t1+1t2+1t3\frac{3}{\frac{1}{t_1} + \frac{1}{t_2}+\frac{1}{t_3}}

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?

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 = a1+a2+a3+.+ann\frac{a_{1}+a_{2}+a_{3}+….+a_{n}}{n}

Geometric mean = a1.a2.a3.ann\sqrt[n]{a_{1}.a_{2}.a_{3}….a_{n}}

Harmonic mean = n1a1+1a2+1a3++1an\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+…+\frac{1}{a_{n}}}

 

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

Relationship: G=A×HG = \sqrt{A\times H}

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 = n1x1+1x2+1x3+.1xn\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+….\frac{1}{x_n}}

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, H=21a+1bi.e.,H=2ab(a+b)H=\frac{2}{\frac{1}{a}+\frac{1}{b}}\,\,\,i.e.,\,\,\,H=\frac{2ab}{(a+b)}

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.

Single HM of n Positive Numbers:

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

H=n(1a1+1a2+1a3++1an)H=\frac{n}{\left(\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+…+\frac{1}{{{a}_{n}}} \right)}

Points to Remember :

1. HM of a, b, c is 31a+1b+1cor3abcab+bc+ca\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\,\,or\,\,\frac{3abc}{ab+bc+ca}

2. The AM between two numbers a and b is a+b2\frac{a+b}{2}.

It does not follow that HM between the same numbers is 2a+b.\frac{2}{a+b}.The HM is the reciprocals of 1a+1b2i.e.,2ab(a+b).\frac{\frac{1}{a}+\frac{1}{b}}{2}i.e.,\,\,\frac{2ab}{(a+b)}.

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, =1(n+2)thtermofcorrespondingAP=\frac{1}{(n+2)th\,term\,of\,corresponding\,AP} =11a+(n+21)D=\frac{1}{\frac{1}{a}+(n+2-1)D}

This implies, D = (1/b – 1/a)/(n+1)

Therefore, 1H1=1a+D,1H2=1a+2D,.,1Hn=1a+nD\frac{1}{H_1} = \frac{1}{a} + D, \frac{1}{H_2} = \frac{1}{a} + 2D, …., \frac{1}{H_n} = \frac{1}{a} + nD 1H1=1a+D,1H2=1a+2D,,1Hn=1a+nD\frac{1}{{{H}_{1}}}=\frac{1}{a}+D,\frac{1}{{{H}_{2}}}=\frac{1}{a}+2D,…,\frac{1}{{{H}_{n}}}=\frac{1}{a}+nD 1H1=1a+(ab)ab(n+1),1H2=1a+2(ab)ab(n+1),..,1Hn\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}}} =1a+n(ab)ab(n+1)=\,\,\,\,\,\frac{1}{a}+\frac{n(a-b)}{ab(n+1)}

Harmonic Mean For Grouped Data

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

H.M = f1+f2+f3++fnf1x1+f2x2+f3x3++fnxn\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}}} = f(fx)\frac{\sum f}{\sum (\frac{f}x{})}

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

HMw=i=1nwii=1nwixiHM_w = \frac{\sum_{i=1}^{n}w_i}{\sum_{i=1}^{n}\frac{w_i}{x_i}}

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: If H be the harmonic mean between x and y, then show that H+xHx+H+yHy=2\frac{H+x}{H-x}+\frac{H+y}{H-y}=2

Solution: We have, H=2xyx+yH=\frac{2xy}{x+y} Hx=2yx+y and Hy=2xx+y\frac{H}{x}=\frac{2y}{x+y}\ and\ \frac{H}{y}=\frac{2x}{x+y}

By componendo and dividendo, we have H+xHx=2y+x+y2yxy=x+3yyx\frac{H+x}{H-x}=\frac{2y+x+y}{2y-x-y}=\frac{x+3y}{y-x} and H+yHy=2x+x+y2xxy=3x+yxy\frac{H+y}{H-y}=\frac{2x+x+y}{2x-x-y}=\frac{3x+y}{x-y} H+xHx+H+yHy=x+3yyx+3x+yxy=x+3y3xyyx=2(yx)(yx)=2\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

Now, H+xHx+H+yHy=2\frac{H+x}{H-x}+\frac{H+y}{H-y}=2 (H+xHx1)=(1H+yHy)2xHx=2yHy\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}

i.e. Hxxy=+Hy+xyH(x+y)=2xyHx-xy=-+Hy+xy\Rightarrow H(x+y)=2xy

i.e. H=2xy(x+y)H=\frac{2xy}{(x+y)}

Which is true as, x, H, 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