Relation Between AM, GM and HM

Generally, Mean means the average of a given set of values in mathematics. In statistics, the mean is one of the measures of central tendency.

The full form of AM, GM, and HM is Arithmetic Mean, Geometric Mean, and Harmonic Mean.

Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) are three different ways of calculating the average or central tendency of a set of numbers. Each mean has its specific application and is used in different contexts to provide meaningful insights from data. The relation between AM, GM, and HM helps us to solve various mathematical problems. AM, GM, and HM relation formula is based on various inequalities and mathematical properties.

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In this chapter, we will discuss AM, GM, and HM, the relationship between AM, GM, and HM, AM, GM, and HM formula, and AM, GM, and HM inequality.

What is Arithmetic Mean?

Arithmetic Mean generally means the average value of a given set of numbers.

It is the easiest way to find the average of a given set of numbers.

Suppose in an exam, you got 70, 78, 85, 90, and 43 in 5 subjects. You need to find your average score. This is when the Arithmetic Mean will be used.

Arithmetic Mean is always greater than Harmonic Mean and greater than or equal to the Geometric Mean.

The formula of Arithmetic Mean

For a set of $n$ positive integers $a_1, a_2, a_3, a_4$,.......$a_n$

Arithmetic mean = $\frac{a_1+a_2+a_3+a_4+.......+a_n}{n}$

For an AP of $n$ terms, AM = $\frac{2a_1+(n-1)d}{n}$ = $\frac{a_1+a_n}{2}$ where n is the number of terms, $a_1$ is the first term of AP and d is the common difference.

First, you have to add all the n observations, then divide that by $n$ to get the Arithmetic Mean.

Example: Arithmetic mean of 70, 78, 85, 90, 43 = $\frac{70+78+ 85+ 90+ 43}{5}=\frac{366}{5}=73.2$

Insertion of n-Arithmetic Mean between a and b

If $A_1, A_2, A_3,......An$ be n number of Arithmetic mean between two numbers $a$ and $b$, then $a, A_1, A_2, A_3,......A_n, b$ will be in Arithmetic progression.

This Arithmetic progression contains $(n + 2)$ terms.

Let d be the common difference here.

$b = (n + 2)th$ term

⇒ $b = a + ((n + 2) -1)d = a + (n+1)d$

$\therefore d =\frac{b-a}{n+1}$

And $A_n= a + (n + 1 - 1)d = a + nd$
Now, we can calculate the values of $A_1, A_2, A_3,......A_n$ by placing the values of $n$.

What is Geometric Mean?

In mathematics, the term mean is defined as the average of the given number of values.

Generally, we get the mean of the numbers by dividing the sum of the numbers by the count of the number.

However, to find the geometric mean, we have to multiply the numbers and then take the root of the corresponding count of numbers.

Geometric mean is very useful for understanding the central tendency of sets of numbers that are multiplicative.

Geometric Mean is always less than or equal to the Arithmetic Mean and greater than or equal to the Geometric Mean.

Formula of Geometric Mean

For a set of $n$ positive integers $a_1, a_2, a_3, a_4$,.......$a_n$

Geometric mean = $\sqrt[n]{a_1× a_2× a_3× a_4× .......× a_n}$

Example: Geometric mean of 2, 4, 8, and 16 = $\sqrt[4]{2×4×8×16}=\sqrt[4]{1024}=5.66$

What is Harmonic Mean?

The Harmonic Mean is the reciprocal of the average of the reciprocals of a given set of numbers.

In simpler terms, you take the average of the reciprocals (1 divided by each number) and then take the reciprocal of that result.

The Harmonic Mean is always less than the Arithmetic Mean and less than or equal to the Geometric Mean.

In general, it is mostly used to find ratios and rates.

Let's take an example.

If you go to a place at a speed of 60 km/hr and return with a speed of 40 km/hr. Then to find the average during this journey Harmonic Mean is used.

Formula of Harmonic Mean

For a set of n positive integers $a_1, a_2, a_3, a_4$,.......$a_n$

Harmonic mean = $\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+.......+\frac{1}{a_n}}$

Using this let's find the average speed during the journey mentioned above.

So, to go 1 km, you will take $\frac{1}{60}$

And to return 1 km, you will take $\frac{1}{40}$

So, average speed = $\frac{\frac{1}{60}+\frac{1}{40}}{2}=\frac{1}{48}$

Finally, we will take the reciprocal of that average i.e., 48 km/hr as the average speed.

Relationship between AM, GM, and HM

For any set of positive real numbers $a_1, a_2, a_3, a_4$,.......$a_n$

Arithmetic Mean × Harmonic Mean = Geometric Mean2

Let’s check if it is true or not.

For two positive integers a and b,

Arithmetic Mean = $\frac{a+b}{2}$

Geometric Mean = $\sqrt{ab}$

Harmonic Mean = $\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}$

Now,

Arithmetic Mean × Harmonic Mean

$=\frac{a+b}{2}×\frac{2ab}{a+b}$

$=ab$

Geometric Mean2

= $(\sqrt{ab})^2$

= $ab$

So, Arithmetic Mean × Harmonic Mean = Geometric Mean2 is true for any set of positive integers.

AM, GM, and HM inequality relationship

For any set of positive real numbers a1, a2, a3, a4, .........an, AM $\geq$ GM $\geq$ HM

Let’s check if it is true.

For two positive integers a and b,

Arithmetic Mean (AM) = $\frac{a+b}{2}$

Geometric Mean (GM) = $\sqrt{ab}$

Harmonic Mean (HM) = $\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}$

Case 1:

$\frac{a+b}{2} \geq \sqrt{ab}$

Squaring both sides, we get,

Squaring both sides, we get,

$⇒ \frac{a^2+2ab+b^2}{4}\geq ab$

$⇒a^2+2ab+b^2 \geq 4ab$

$⇒a^2-2ab+b^2 \geq 0$

$⇒(a-b)^2\geq 0$, which is always true.

So, AM $\geq$ GM

Case 2:

$\sqrt{ab}\geq\frac{2ab}{a+b}$

Squaring both sides, we get,

$ab \geq \frac{4a^2b^2}{a^2+2ab+b^2}$

$⇒a^2+2ab+b^2 \geq4ab$

$⇒a^2-2ab+b^2 \geq 0$

$⇒(a-b)^2\geq 0$, which is always true.

So, GM $\geq$ HM

Hence, from case 1 and case 2, we get,

AM $\geq$ GM $\geq$ HM

So, the inequality holds for any set of positive integers.

Properties of AM, GM, and HM

• For any set of positive real numbers a1, a2, a3, a4, .........an, AM $\geq$ GM $\geq$ HM

• For any set of positive real numbers a1, a2, a3, a4, .........an,

Arithmetic Mean × Harmonic Mean = Geometric Mean2

• A and G are arithmetic and the geometric mean of ‘a’ and ‘b’, two real, positive, and distinct numbers. Then, a and b are the roots of the equation x2 - 2Ax + G2 = 0.

Also, a and b are given by $A \pm \sqrt{(A+G)(A-G)}$

• The arithmetic mean is used for general averaging purposes.

• A geometric mean is used in multiplicative processes like growth rates.

• The harmonic mean is used to find the average speed.

• All numbers must be positive in the dataset for the Geometric mean to be defined.

• The arithmetic mean is changed linearly with each number of a set. If you add or subtract a constant from each number of a set, then the Arithmetic mean will be defined also after subtracting or adding that constant from the initial Arithmetic mean.

Tips and Tricks

• For two positive integers a and b, the Arithmetic mean is $\frac{a+b}{2}$

For three positive integers a, b, and c, the Arithmetic mean is $\frac{a+b+c}{2}$

• For two positive integers a and b, the Geometric mean is $\sqrt{ab}$

For more numbers, the nth root of their product is the Geometric mean.

• For two positive integers a and b, the Harmonic mean is $\frac{2ab}{a+b}$

The harmonic mean of three terms $a, b$ and $c$ is $\frac{3abc}{ab+bc+ca}$

• In an evenly spaced sequence (AP) like 2, 4, 6, 8, …., the Arithmetic mean is the same as the middle number if the number of terms is odd. If the number or terms are even, then the average of two middle numbers is the required Arithmetic mean.

• Harmonic mean can be used in databases of ratios and rates.

Solved Examples

Q1. Find the Arithmetic mean of 5, 10, and 15.

1. 15

2. 10

3. 20

4. 25

Answer:

Sum of the numbers = 5 + 10 + 15 = 30

Count of numbers = 3

$\therefore$ Arithmetic mean = $\frac{30}{3}=10$

Hence, the correct answer is 10.

Q2. Find the Geometric mean of 4, 8, and 32. (to the nearest integer)

1. 10

2. 8

3. 12

4. 15

Answer:

Product of the numbers = 4 × 8 × 32 =1024

Count of numbers = 3

$\therefore$ Geometric mean = $\sqrt[3]{1024} \approx 10$

Hence, the correct answer is 10.

Q3. Find the Harmonic mean of 6 and 12.

1. 8

2. 10

3. 12

4. 6

Answer:

Harmonic mean = $\frac{2×6×12}{6+12}=8$

Hence, the correct answer is 8.

Q4. If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and (a + n) = 33, then the value of n is:

1. 21

2. 22

3. 23

4. 24

Answer:

Let the common difference of the AP obtained be d.

So, first AM = a + d

Last mean = 100 - d

$\mathrm{Their \: ratio =\frac{a+d}{100-d}=\frac{1}{7}\\\Rightarrow 7 a+8 d=100}---(1)$

$\text { Also } \mathrm{100=a+(n+2-1) d} $ (Since total terms = n + 2)

$\mathrm{\Rightarrow 100=a+(n+1) d} $

$\mathrm{\Rightarrow 100=33-n+(n+1) d \quad(\text { given } a+n=33)}$

$\mathrm{\Rightarrow d=\frac{n+67}{n+1} }$

Using equation 1, we get,

$\mathrm{ 7(33-n)+\frac{8(n+67)}{n+1}=100} $

$\mathrm{\Rightarrow 7 n^{2}-132 n-667=0}$

$\mathrm{\therefore n=23 }$

Hence, the correct answer is 23.

Q5. n Arithmetic means are inserted between 3 and –13. If the 3rd AM is –3, then determine the value of n.

1. 7

2. 6

3. 5

4. 8

Answer:

$3, A_1, A_2, A_3,.......A_n,-13$ are in Arithmetic progression.

So, $-13=3+(n+2-1)d$

$\Rightarrow -16=(n+1)d---(1)$

Now, $A_3=-3$

$\Rightarrow 3+3.d=-3$

$\therefore d=-2$

Putting the value of d in equation 1, we get,

$-16=(n+1)(-2)$

$\therefore n=7$

Hence, the correct answer is 7.

Q6. If the arithmetic mean and geometric mean of the pth and qth terms of the sequence –16, 8, –4, 2..... satisfy the equation 4x2 -9x + 5 = 0, then p+q is equal to:

1. 8

2. 10

3. 12

4. 14

Answer:

Given sequence: –16, 8, –4, 2.....

a = –16, r = $-\frac12$

$T_{p}=-16\left(-\frac{1}{2}\right)^{p-1}=-16 \times-2 \times\left(-\frac{1}{2}\right)^{p}=32 \times\left(-\frac{1}{2}\right)^{p}$

$T_{q}=-16\left(-\frac{1}{2}\right)^{q-1}=32 \times\left(-\frac{1}{2}\right)^{q}$

$\begin{aligned} &\frac{T_{p}+T_{q}}{2}=\text { A.M. and } \sqrt{T_{p}.T_{q}}=\text{G . M.}\\ &\text { A.M. and G.M. are the roots of } 4 x^{2}-9 x+5=0 \end{aligned}$

$\text { A.M. }=\frac{32 \times\left(\frac{-1}{2}\right)^{\mathrm{p}}+32 \times\left(\frac{-1}{2}\right)^{\mathrm{q}}}{2}=16\left[\left(\frac{-1}{2}\right)^{\mathrm{p}}+\left(-\frac{1}{2}\right)^{\mathrm{q}}\right] $

$\text { G.M. }=\sqrt{32 \times\left(\frac{-1}{2}\right)^{\mathrm{P}}\times32 \times\left(\frac{-1}{2}\right)^{\mathrm{q}}}=32 \sqrt{\left(\frac{-1}{{2}}\right)^{\mathrm{p}+\mathrm{q}}}$

$4 x^{2}-9 x+5=0$

$\Rightarrow x=1, \frac{5}{4}$

$32 \sqrt{\left(\frac{-1}{2}\right)^{p+q}}=1$

$\therefore p+q=10$

Hence, the correct answer is 10.

Q7. If the A.M. of two positive numbers a and b is 5 and their G.M. is 3, then numbers a and b are:

1. 2 and 8

2. 3 and 7

3. 1 and 9

4. 1 and 10

Answer:

Given that AM = 5

So, $\frac{a+b}{2}=5$

$\Rightarrow a+b=10 ----(1)$

GM = 3

So, $\sqrt{ab}=3$

$\Rightarrow ab = 9 -----(2)$

Now, $(a-b)^2 = (a+b)^2 - 4ab = 100 - 36 = 64$

$ \Rightarrow (a-b)=\pm 8 ----(3)$

From equation 1 and 3, we get,

(a = 9, b = 1) or (a = 1, b = 9)

Hence, the correct answer is option 3.

Q8. The sum of two numbers is 15 and their geometric mean is 20% lower than their arithmetic mean.

Find the numbers.

1. 11 and 4

2. 12 and 3

3. 13 and 2

4. 10 and 5

Answer:

We will solve this problem by checking the options.

Option 1: Here AM = $\frac{15}{2}=7.5$
GM = $\sqrt{11×4}=6.63$, which is not 20% less than 7.5

Option 2: Here AM = $\frac{15}{2}=7.5$
GM = $\sqrt{12×3}=6$, which is 20% less than 7.5

Option 3: Here AM = $\frac{15}{2}=7.5$
GM = $\sqrt{13×2}=5.099$, which is not 20% less than 7.5

Option 4: Here AM = $\frac{15}{2}=7.5$
GM = $\sqrt{10×5}=7.07$, which is not 20% less than 7.5

Hence, the correct answer is option 2.

Q9. If HM : GM = 4 : 5 for two positive numbers, then the ratio of the numbers is:

1. 4 : 1

2. 3 : 2

3. 3 : 8

4. 3 : 4

Answer:

Let the numbers be a and b.

$\frac{HM}{GM} = \frac{4}{5}$

$ \Rightarrow \frac{2ab}{(a+b)\sqrt{ab}} = \frac{4}{5}$

$\Rightarrow \frac{2\sqrt{ab}}{(a+b)}=\frac{4}{5}$

$\Rightarrow 5\sqrt{ab}= 2a + 2b$

$\Rightarrow 2a - 5\sqrt{ab} + 2b = 0$

Dividing by b, we get,

$\Rightarrow 2(\frac ab) - 5\sqrt{\frac ab} + 2 = 0$

Let $\sqrt{\frac ab}$ = t

2t2 – 5t + 2 = 0

t = 2, $\frac12$

Now, putting the value of t, we get,

When, t = 2,

$\sqrt{\frac ab}=2$

$⇒a:b=4:1$

When, t = $\frac12$,

$\sqrt{\frac ab}=\frac12$

$⇒a:b=1:4$

Hence, the correct answer is the option 1.

Q10. Find out the AM, GM, and HM of the series 2, 5, 8, 9.

Answer:

Arithmetic mean = $\frac{2+5+8+9}{4}=6$

Geometric mean = $ \sqrt[4]{2\times 5\times 8\times 9}=5.18$

We know, Arithmetic Mean × Harmonic Mean = Geometric Mean2

So, Harmonic mean

= $\frac{\text{Geometric Mean × Geometric Mean}}{\text{Arithmetic Mean}}$

= $\frac{5.18×5.18}{6}$

= $4.47$