📢 Relation Between AM, GM & HM – Class 11 Mathematics 📚
🚀 Master Arithmetic Mean (AM), Geometric Mean (GM), & Harmonic Mean (HM) for Board & JEE Exams! 🚀
Understanding the relation between Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) is essential in Class 11 Mathematics, particularly in Sequences & Series. These means are widely used in algebra, statistics, physics, and real-world applications, making them crucial for CBSE Board Exams, JEE Mains, and Advanced.
Table of Contents
🔢 AM, GM, HM Formulas
Before we relate the three means in Statistics, which are Arithmetic Mean, Geometric Mean and Harmonic Mean, let us understand them better.
Arithmetic Mean
Arithmetic mean represents a number that is achieved by dividing the sum of the values of a set by the number of values in the set. If a1, a2, a3,….,an, is a number of group of values or the Arithmetic Progression, then;
AM=(a1+a2+a3+….,+an)/n
Geometric Mean
The Geometric Mean for a given number of values containing n observations is the nth root of the product of the values.
GM = n√(a1a2a3….an)
Or
GM = (a1a2a3….an)1/n
Harmonic Mean
HM is defined as the reciprocal of the arithmetic mean of the given data values. It is represented as:
HM = n/[(1/a1) + (1/a2) + (1/a3) + ….+ (1/an)]
🔢 What is the Relation between AM, GM and HM?
The relationship between AM, GM and HM is given by:
| AM x HM = GM2 |
Now let us understand how this relation is derived;
First, consider a, AM, b is an Arithmetic Progression.
Now the common difference of Arithmetic Progression will be;
AM – a = b – AM
a + b = 2 AM …………..(1)
Secondly, let a, GM, b is a Geometric Progression. Then, the common ratio of this GP is;
GM/a = b/GM
ab = GM2……………(2)
Third, is the case of harmonic progression, a, HM, b, where the reciprocals of each term will form an arithmetic progression, such as:
1/a, 1/HM, 1/b is an AP.
Now the common difference of the above AP is;
1/HM – 1/a = 1/b – 1/HM
2/HM = 1/b + 1/a
2/HM = (a + b)/ab ………….(3)
Substituting eq. 1 and eq.2 in eq. 3 we get;
2/HM = 2AM/GM2
GM2 = AM x HM
Hence, this is the relation between Arithmetic, Geometric and Harmonic mean.
Relationship between AM, GM and HM
- For any set of unequal positive numbers, the relationship between AM, GM, and HM is expressed as: AM ≥ GM ≥ HM.
- If the values in the set data are equal, then the three averages would also be equal.
AM–GM Inequality Relationship
For ‘n’ positive numbers a1, a2, a3, a4, … an
Arithmetic Mean: A.M = (a1+ a2+ a3+ a4+………. +an) / n
Geometric Mean: G.M = (a1 a2 a3 a4 ………. an) 1/n
A.M ≥ G.M
(a1+ a2+ a3+ a4+………. +an) / n ≥ (a1 a2 a3 a4 ………. an) 1/n
AM–GM Inequality Relationship Proof
Statement: For any n positive numbers a1, a2, … an Arithmetic Mean is always greater than equal to Geometric Mean. A.M ≥ G.M
Proof:
Hence we conclude by above proof that for all positive Numbers, A.M ≥ G.M
For two numbers,
A.M – G.M = (a+b)/2 – √(ab)
A.M – G.M = ½ (a+b – 2 √(ab) )
A.M – G.M = ½ (√a – √b)2
We know that square of any number is positive,( i.e ≥0) Hence,
A.M – G.M ≥0
A.M ≥ G.M
🔍 Applications of AM, GM & HM
📌 Statistics & Data Science – Mean calculations in datasets
📌 Finance & Economics – Interest rates, investment returns
📌 Physics & Engineering – Harmonic motion, resistance in circuits
📌 Chemistry & Biology – Growth rates, molecular concentrations
📌 Mathematics & Optimization – Inequality proofs, algorithm efficiency
Prove the Inequality : AM > GM > HM
Example 1:
For the two data points 10, 12, find the relationship between AM, GM, HM, and arrange them in ascending order.
Solution:
The given data points are 10, 12
Arithmetic Mean (AM) = (10 + 12)/2 = 22/2 = 11
Geometric Mean(GM) = √10×12 = √120 = 10.94
Harmonic Mean (HM) = 2(10)(12)/(10+12) = 10.9
From the above values we have 11 > 10.94 > 10.9
AM > GM > HM
Therefore, we have AM > GM > HM.
Example 2:
Find the value of geometric mean GM, if the arithmetic mean AM is 7, and harmonic mean HM is 48/7.
Solution:
Arithmetic Mean (AM) = 7
Harmonic Mean(HM) = 48/7
Geometric Mean(GM) = ?
The relation between AM, GM, HM can be computed using the following formula.
AM × HM = GM2
7 × 48/7 = GM2
GM2 = 48
GM = √48
GM = 4√3
Therefore the geometric mean of the two numbers is 4√3.
Example 3:
If five times the geometric mean of two numbers ‘a’ and ‘b’ is equal to the arithmetic mean of those two numbers such that a > b > 0, then compute the value of a + ba – b.
Solution:
Arithmetic mean of the two numbers is calculated as
\[AM = \frac{(a+b)}{2}\]
Geometric mean of the two numbers a and b is
\[GM = \sqrt{ab}\]
It is given in the question that Arithmetic mean = 5 times the geometric mean
AM = 5 GM
\[\frac{a+b}{2} = 5\sqrt{ab}\]
a + b = 10 √ab
(a + b)2 = 100 ab
(a – b)2 = (a + b)2 – 4 ab
(a – b)2 = 100 ab – 4 ab
\[(a-b)^{2} = 96ab\]
a-b = \[\sqrt{96ab}\]
\[\frac{a+b}{a-b} = \sqrt{\left ( \frac{100ab}{96ab} \right )} = \sqrt{\left ( \frac{25}{24} \right )} = 1.021\]
Example 4:
Find the harmonic mean of two numbers a and b, if the arithmetic mean is 25 and the geometric mean is 10 provided that a>b>0.
Solution:
Given, A.M. = 25 and G.M. = 10
The relationship between AM, GM, and HM is, G.M.2=A.M.×H.M.
102=25×H.M.
100=25×H.M.
H.M.=100/25
H.M. = 4
Example 5:
If x>0, then Prove That: x+ (1/x) ≥ 2
Solution:
Since x>0 we can apply A.M-G.M Inequality here,
A.M ≥ G.M
(x+1/x) / 2 ≥ (x . 1/x) ½
(x+1/x) /2 ≥ 1
x + (1/x) ≥ 2 (Proved)
Example 6:
If x,y>0,then Prove That: x2+y2 ≥ 2xy
Solution:
Since x,y>0 we can apply A.M-G.M Inequality here,
A.M ≥ G.M
For two variables x and y
(x+y) /2 ≥ (xy)1/2
squaring both sides:
(x+y)2 /4 ≥ xy
x2+ y2 ≥ 2xy
Example 7:
If x,y,z ≤ 0,then can we Prove That: (x+y)(y+z)(z+x) ≥ 8xyz through A.M-G.M Inequality
Solution:
Since x, y, z ≤ 0
So, we can’t apply A.M-G.M Inequality here,
Example 8:
If a,b,c ∈ R+, such that a+b+c = 3, find the maximum value of abc.
Solution:
Since a,b,c >0 we can apply A.M-G.M Inequality here.
We need to find the value of the product of a,b,c i.e abc.
Applying A.M ≥ G.M,
(a+b+c)/3 ≥ 3√(abc)
3/3 ≥ (abc)1/3
1 ≥ (abc)1/3
cubing both sides, 13 ≥ (abc)
so, abc ≤1
Hence the maximum value of abc = 1
Example 9:
Find the harmonic mean of two numbers a and b, if their arithmetic mean is 16 and geometric mean is 8 provided that a > b > 0. (Hint: Use relation between AM GM HM formula).
Solution:
Given: AM = 16 and GM = 8
The relation between AM GM HM is given as:
\[AM \times HM = GM^{2}\]
\[16 \times HM = 8^{2} \]
\[16 \times HM = 64 \]
HM = 64/16 = 4
To find the numbers:
Arithmetic mean is given as
\[AM = \frac{(a+b)}{2}\]
\[16 = \frac{(a+b)}{2}\]
a + b = 32
a = 32 – b
Geometric mean is given as
GM = √ab
8 = √ab
64 = ab
64 = (32 – b) b
64 = 32b – b2
b2 – 32 b + 64 = 0
Formulas
\[AM = \frac{(a+b)}{2}\]
\[GM = \sqrt{a+b}\]
\[HM = \frac{2ab}{a+b}\]
- If zero is one of the terms of a sequence, its geometric mean is zero and the harmonic mean is infinity.
Solved Examples for JEE Exam
Example 1: Prove that
\(\begin{array}{l}(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )\ge g\alpha \beta \gamma\end{array} \)
Solution: Applying AM > GM
\(\begin{array}{l}\frac{\alpha +\beta +\gamma }{3}\ge {{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(1)\end{array} \)
\(\begin{array}{l}\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3}\ge {{\left( {{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}} \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(2)\end{array} \)
On multiplying equations (1) and (2),
We get
\(\begin{array}{l}\frac{\alpha +\beta +\gamma \left( \alpha \beta +\beta \gamma +\gamma \alpha \right)}{g}\ge {{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\,{{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\ge \left( \alpha \beta \gamma \right)\end{array} \)
Therefore,
\(\begin{array}{l} (\alpha +\beta +\gamma )\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)\ge g\alpha \beta \gamma\end{array} \)
Hence proved.
Example 2: Insert 3 GMs between 1 and 16
Solution: Let the terms be 1, g1, g2, g3, 16
\(\begin{array}{l}r={{\left( \frac{16}{1} \right)}^{\frac{1}{3+1}}}\end{array} \)
\(\begin{array}{l}r={{\left( 16 \right)}^{1/4}}=2\end{array} \)
Three geometric means between 1 and 16 are 2, 4, 8.
Example 3: If P, Q, R are positive real numbers such that P + Q + R = g, find the maximum value of P2Q3R4.
Solution: P + Q + R = g
Applying AM > GM
\(\begin{array}{l}2\left( \frac{P}{2} \right)+3\left( \frac{Q}{3} \right)+4\left( \frac{R}{4} \right)=g\end{array} \)
\(\begin{array}{l}\Rightarrow \left( {{\left( \frac{P}{2} \right)}^{2}}.{{\left( \frac{Q}{3} \right)}^{3}}.{{\left( \frac{R}{4} \right)}^{4}} \right)\le \frac{g}{3}\end{array} \)
3g, 22, 33, 44 > P2Q2R4
\(\begin{array}{l}{{3}^{12}}\cdot {{2}^{10}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}}\end{array} \)
\(\begin{array}{l}{{6}^{10}}\cdot {{3}^{2}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}}\end{array} \)
Example 4: Evaluate 12 + 22 + 32 + 42 + 52.
Solution: Above expression can be written as
\(\begin{array}{l}\sum_{m=1}^5 m^2 \end{array} \)
We know, sum of squares of first five natural numbers = 1/6 {n(n+1)(2n+1)}
Here, n = 5.
Therefore,
\(\begin{array}{l}\sum_{m=1}^5 m^2 \end{array} \)
= 5(5+1)(2.5+1)/6 = 55
Example 5: Prove (a + b + c) (ab + bc + ca) > 9abc.
Solution:
\(\begin{array}{l}\frac{a+b+c}{3}>(abc)^{1/3}\end{array} \)
and
\(\begin{array}{l}\frac{ab+bc+ca}{3}> (ab.bc.ca)^{\frac{1}{3}}\end{array} \)
[Using AM > GM]
Multiplying the above results, we get
\(\begin{array}{l}\frac{a+b+c}{3} \times \frac{ab+bc+ca}{3} > (abc)^{1/3}(ab.bc.ca)^{1/3}\end{array} \)
= (a3b3c3)1/3
= abc
⇒ (a + b + c)(ab + bc + ca) > 9abc.
Hence proved.
Example 5: Consider the function
\(\begin{array}{l}f(x,y,z)={\frac {x}{y}}+{\sqrt {{\frac {y}{z}}}}+{\sqrt[ {3}]{{\frac {z}{x}}}}\end{array} \)
for all positive real numbers x, y and z. Find the minimal value of the function.
Solution:
\(\begin{array}{l}{\begin{aligned}f(x,y,z)&=6\cdot {\frac {{\frac {x}{y}}+{\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}+{\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}}{6}}\\&=6\cdot {\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\end{aligned}}\\ with \ {\displaystyle x_{1}={\frac {x}{y}},\qquad x_{2}=x_{3}={\frac {1}{2}}{\sqrt {\frac {y}{z}}},\qquad x_{4}=x_{5}=x_{6}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}\end{array} \)
Applying the AM–GM inequality for n = 6, we get
\(\begin{array}{l}{\begin{aligned}f(x,y,z)&\geq 6\cdot {\sqrt[ {6}]{{\frac {x}{y}}\cdot {\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}\cdot {\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}}}\\&=6\cdot {\sqrt[ {6}]{{\frac {1}{2\cdot 2\cdot 3\cdot 3\cdot 3}}{\frac {x}{y}}{\frac {y}{z}}{\frac {z}{x}}}}\\&=2^{{2/3}}\cdot 3^{{1/2}}.\end{aligned}}\end{array} \)
Further, we know that the two sides are equal exactly when all the terms of the mean are equal.
\(\begin{array}{l}f(x,y,z)=2^{{2/3}}\cdot 3^{{1/2}}\text{when}\ {\frac {x}{y}}={\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}={\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}.\end{array} \)
All the points (x, y, z), satisfying these conditions lie on a half-line starting at the origin and are given by,
\(\begin{array}{l}{\displaystyle (x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\text with \ t \ >0.}\end{array} \)
Practice Problems
- Find the arithmetic mean, geometric mean, and harmonic mean of the numbers 5, 10, and 15.
- Given two numbers a and b, where A.M.=12 and G.M. =8, find HM.
- Show that G.M.2=A.M.×H.M. for 9, 25, 35.
- Find the arithmetic mean, geometric mean, and harmonic mean of the numbers 2, 8, 32, and 128.
- A sequence of numbers follows a geometric progression with the first term a = 2 and common ratio r = 3. If there are 5 terms in the sequence, calculate the arithmetic mean, geometric mean, and harmonic mean of the sequence.
- Given a dataset where the values follow a harmonic progression: 1, 1/2, 1/3, 1/4, and 1/5, calculate the arithmetic mean, geometric mean, and harmonic mean.
Relationship between AM, GM and HM – FAQs
What is the primary difference between Arithmetic Mean and Geometric Mean?
The Arithmetic Mean is the sum of all observations divided by their number, providing a simple average. The Geometric Mean is the nth root of the product of n values, reducing the impact of extreme values.
When should the Harmonic Mean be used?
The Harmonic Mean is useful in situations requiring an average rate, such as speeds or rates of work.
Can the Geometric Mean be used for negative numbers?
No, the Geometric Mean is only defined for positive numbers.
What happens to AM, GM, and HM if all values in a dataset are equal?
If all values are equal, then AM, GM, and HM will all be equal to that value.
How is the relationship AM>GM>HM significant?
It shows that for any set of unequal positive numbers, the Arithmetic Mean will always be greater than the Geometric Mean, which in turn will be greater than the Harmonic Mean.
📚 Why Choose Neeraj Anand’s Book?
✔️ Concept-Based Learning – Clear explanations & step-by-step proofs
✔️ Exam-Oriented Approach – Focused on CBSE Board & JEE syllabus
✔️ Practice Questions – Solved examples and exercises for mastery
📥 Download the PDF & Start Learning Today!
📕 Author: Neeraj Anand
🏛 Publisher: ANAND TECHNICAL PUBLISHERS
🏫 Available at: ANAND CLASSES
💯 Strengthen Your Concepts in AM, GM & HM & Excel in Your Exams!