PROBLEMS ON HARMONIC MEAN
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Harmonic mean(H) is one of the measures of central tendency and also based on arithmetic mean and geometric mean. It is useful for quantitative data. It is also a kind of average.
Definition :
This is nothing but the reciprocal of the arithmetic mean of the reciprocals of the given numbers.
In other words to find the H of the given numbers, we have to divide n (total number of given numbers) by the sum of the reciprocals of the given numbers.
The series x1, x2, x3, ............xn is said to be in Harmonic progression when their reciprocals 1/x1, 1/x2, 1/x3, ...........1/xn are in arithmetic progression.
- When three numbers are in HP the middle number is called the Harmonic mean between the other two numbers.
- If a, H, b are in H.P then H is Harmonic mean of a and b.
- Also 1/a, 1/H, 1/b are in H.P
- H = 2ab/(a + b)
Example 1 :
Find the H of 3, 4, 5, 6, 7 and 8
Solution :
The total number of values = 6
Let us find H using the formula
H = n /(1/a₁ + 1/a₂ +.....+1/aₓ)
= 6/(1/3+1/4+1/5+1/6+1/7+1/8)
= 6/(0.333+0.25+0.20+0.166+0.142+0.125)
= 6/1.216
= 4.93
Example 2 :
Find the H of 1, 2, 5, 7, 9
Solution :
The total number of values = 5
Let us find H using the formula
H = n /(1/a₁ + 1/a₂ +.....+1/aₓ)
= 6/(1/1+1/2+1/5+1/7+1/9)
= 6/(1 + 0.5 + 0.2 + 0. 14 + 0.11)
= 6/1.95
= 3.07
Harmonic mean of two numbers
Example 3 :
Find the H of two numbers 50 and 30.
Solution :
Instead of x1 and x2 we are having 50 and 30.
= 2(50) (30) / (50 + 30 )
= (100 x 30)/80
= 3000/80
= 37.5
Example 4 :
Find the H of two numbers 12 and 15.
Solution :
Instead of x1 and x2 we are having 12 and 15.
= 2(12) (15) / (12 + 15 )
= (24 x 15)/27
= 360/27
= 13.3
Example 5 :
Find the H of two numbers 12 and 15.
Solution :
Instead of x1 and x2 we are having 12 and 15.
= 2(12) (15) / (12 + 15 )
= (24 x 15)/27
= 360/27
= 13.3
Example 6 :
Find all three means for a = 36 and b = 64
Solution :
a = 36 and b = 64
= 2ab/(a + b)
= 2(36)(64) / (36 + 64)
= 4608/100
= 46.08
Example 7 :
Insert 3 harmonic means between 1/3 and 1/9.
Solution :
Harmonic mean of two numbers = 2ab/(a + b)
c, d and e are harmonic means.
First harmonic mean :
a = 1/3 and b = 1/9
= 2(1/3)(1/9) / (1/3 + 1/9)
= (2/27)/(3+1)/9
= (2/27) / (4/9)
= (2/27) x (9/4)
= 1/6
Second harmonic mean :
= 2(1/3)(1/6) / (1/3 + 1/6)
= (1/9)/(3/6)
= (1/9) x (6/3)
= 2/9
Third harmonic mean :
= 2(2/9)(1/9) / (2/9 + 1/9)
= (4/81)/(3/9)
= (4/81) x (9/3)
= 4/27
So, the values of c, d and e are 1/6, 2/9 and 4/27.
Example 8 :
Calculate harmonic mean from the following data.
3, 13, 11, 15, 5, 4, 2
Solution :
|
x 3 13 11 15 5 4 2 |
1/x 1/3 = 0.333... 1/13 = 0.0769 1/11 = 0.0909 1/15 = 0.0667 1/5 = 0.2 1/4 = 0.25 1/2 = 0.5 |
Σ(1/x)
= 0.333... + 0.0769 + 0.0909 + 0.6666 + 0.2 + 0.25 + 0.5
= 1.515
Harmonic mean = n/Σ(1/x)
= 7/1.515
= 4.62
So, the required harmonic mean is 4.62.
Example 9 :
Calculate harmonic mean from the following data.
10, 50, 30, 20, 10, 20, 70, 30
Solution :
|
x 10 50 30 20 10 20 70 30 |
1/x 1/10 = 0.1 1/50 = 0.02 1/30 = 0.033 1/20 = 0.05 1/10 = 0.1 1/20 = 0.05 1/70 = 0.01 1/30 = 0.033 |
Σ(1/x)
= 0.1 + 0.02 + 0.033 + 0.05 + 0.1 + 0.05 + 0.01 + 0.033
= 0.396
Harmonic mean = n/Σ(1/x)
= 8/0.396
= 20.20
So, the required harmonic mean is 20.20
Example 10 :
Calculate Harmonic mean from the following data
73, 70, 71, 73, 68, 67, 69, 72, 76, 71
|
x 73 70 71 73 68 67 69 72 76 71 |
1/x 1/73 = 0.014 1/70 = 0.014 1/71 = 0.014 1/73 = 0.014 1/68 = 0.015 1/67 = 0.015 1/69 = 0.014 1/72 = 0.013 1/76 = 0.013 1/71 = 0.014 |
Σ(1/x) =
0.014 + 0.014 + 0.014 + 0.014 + 0.015 + 0.015 + 0.014 + 0.013 + 0.013 + 0.014
= 0.14
Harmonic mean = n/Σ(1/x)
= 10/0.14
= 71.42
So, the required harmonic mean is 71.42.
Example 11 :
Find the 29th term of the series.
1/4, 1/7, 1/11, 1/14, ........
Solution :
1/4, 1/7, 1/11, 1/14, ........ are in H.P
4, 7, 11, 14,........... are in A.P
Here a = 4, d = 7 - 4
d = 3
29th term :
a29 = a + 28d
= 4 + 28(3)
= 4 + 84
= 88
88 is the 29th term of the arithmetic progression.
1/88 is the 29th term of harmonic progression.
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