Harmonic Mean Problems

Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) are all measures of central tendency. If all the observations of the series are constant ‘K’, the mean will be K’. The same property applies for all the means, be it arithmetic, geometric, or harmonic. 

If we take the deviation in the series from the mean, the sum of deviations from the mean will be zero. If we change the origin, we will observe the same change consequently in the mean. Similarly, if we change the scale, the same scale change will be seen eventually in the mean. 

We will study all the concepts below, including the mean, its calculation and the relationship between Arithmetic Mean, Geometric Mean and Harmonic Mean.

Harmonic Mean Problems

We will discuss the following categories of harmonic mean problems.

1. All the observations are constant (same), means will be the same constant value

2. Problems on Harmonic Mean of given Numbers

3. Problems on Insertion of Harmonic Mean between two given numbers

4. Problems on Relationship between AM, GM and HM

1. All the observations are constant(same): AM, GM and HM will also be the same constant.

Question: X = 5, 5, 5, 5, 5

Solution:

AM = 5

GM = 5

HM = 5

2. Problems on Harmonic Mean of given Numbers

Question:

Calculate the Harmonic Mean for given observations: 2, 4 and 6.

Solution:

Given Observations are:  2,4,6

No. of observations (n) = 3

Step 1:

The sum of the reciprocal of the observations will be:

1/x1 + 1/x2 + 1/x3+…………………………..+ 1/xn         …………………….(i)

Putting the values in the equation (i) above, we will get

= 1/2 + 1/4  + 1/6

Step 2:

The Arithmetic Mean of reciprocal of observations can be calculated as follows:

(1/x1 + 1/x2 + 1/x3+…………………………..+ 1/xn ) / n        ………………………….(ii)

Putting the values in the equation (ii) above, we will get

= (1/2 + 1/4 + 1/6) / 3

Step 3:

Now, the reciprocal of Arithmetic Mean of the reciprocal of the observations will be simply the reciprocal of what we just calculated above in step (ii):

n / (1/x1 + 1/x2 + 1/x3+…………………………..+ 1/xn ) ………………………(iii)

Putting the values in equation (iii) above, we will get

= 3 / (1/3 + 1/4 + 1/5)

= 3 / (0.5 + 0.25 + 0.167)

Solving the above equation, we will get

Harmonic Mean = 3.27

3. Problems on Insertion of Harmonic Mean between two given numbers:

Question: Insert three Harmonic Means between two numbers 16 and 16/5.

Solution:

Given Harmonic Progression (H.P.) will be as below:

16, H1, H2, H3, 16/5

a1 = 16/1 (first term)

a5 = 16/5 (fifth term)

We know that:  A1,  A2,  A3,………………………….An is an A.P.

Here

A1 = 1/16 (Reciprocal of a1)

A5 = 5/16 (Reciprocal of a6)

Hence, we can calculate d from here:

a+4d = 5/16

1/16 + 4d = 5/16

d = 1/16

Now, as we have the value of d, we can calculate all the missing values as follows:

A2 = (1+1)/16 = 2/16

Hence H1 = 16/2=8

A3 = 3/16

H2 = 16/3

A4 = 4/16

H3 = 4

4. Problems on Relationship between AM, GM, and HM

For given positive numbers, the given result will always hold true as below:

AM >= GM >= HM

For two numbers, we can interpret the result as:

[ (a+b)/2 >=  √ab >= 2ab / (a+b) ]

We can derive the result for ‘n’ numbers in the same pattern.

Also, AM*HM = GM2 

Question: Find AM, HM and GM for 5, 15 and find the relationship between them.

Solution:

AM = Sum of observations/ No, of observations = (5+15)/2 = 10

GM = (x1*x2)1/2 = (5*15)1/2 = 8.66

HM = n / (1/x1 + 1/x2 + 1/x3+…………………………..+ 1/xn ) = 2/(1/5+1/15) = 7.5

AM*HM = 10*7.5 = 75

GM2 = 8.66*8.66 = 75

Hence AM*HM = GM2

Conclusion

The harmonic mean is based on the reciprocal of numbers averaged. We can define it as the reciprocal of the arithmetic mean of the reciprocal of the individual observations. 

Advantages of harmonic mean include that we can base its value on every item of the series. It tends itself to algebraic manipulations. In problems relating to time and rates, it gives better results than other averages. However, it is not easily understood and is difficult to compute compared to other means. 

Moreover, we cannot compute its value when there are both positive and negative items in a series or when one or more items are zero. We solved different practice problems in this article. We hope this study material will help you get better concept clarity on solving harmonic mean problems stepwise and simple manner.