Relationship between Arithmetic Mean & Geometric Mean-Arithmetic Mean ≥ Geometric Mean, Solved Examples

📢 Relationship Between Arithmetic Mean (AM) and Geometric Mean (GM) – Class 11 Mathematics 📚

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The relationship between Arithmetic Mean (AM) and Geometric Mean (GM) is a crucial topic in Class 11 Mathematics, playing a vital role in Inequalities, Sequences & Series, and Competitive Exams like JEE Mains & Advanced. Understanding this concept helps students develop strong problem-solving skills for advanced mathematical applications.

Relationship between Arithmetic Mean and Geometric Mean

Relation 1:

Arithmetic Mean ≥ Geometric Mean

AM = A = (a + b)/2 . . . . . . . (1)

And, GM = G = √(ab)  . . . . . . . . . . (2)

Now, Equation (1) – Equation (2), we get

\(\begin{array}{l}\mathbf{A – G = \frac{a\;+\;b}{2}\;-\;\sqrt{ab}}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{\frac{a\;+\;b\;-\;2\;\sqrt{ab}}{2}\;=\;\frac{\left ( \;\sqrt{a}\;-\;\sqrt{b}\; \right )^{2}}{2}\;\geq\; 0}\end{array} \)

i.e. A – G ≥ 0

This implies that A ≥ G.

Note : ✅ If all numbers in the set are equal, then GM = AM.

Relation 2:

The quadratic equation whose roots are a and b can be written as x² – 2Ax + G² = 0. (Where A and G are arithmetic and geometric mean between a and b, respectively.)

If a and b are the roots of the quadratic equation, then x² – (a + b)x + ab = 0

Since, a + b = 2A and ab = G²

Therefore, the above equation becomes x² – 2Ax + G² = 0

Relation 3:

If A and G are arithmetic and geometric mean between two numbers a and b, respectively, then

\(\begin{array}{l}\mathbf{a\;=\;A\;+\;\sqrt{A^{2}\;-\;G^{2}}\;\;and\;\;b\;=\;A\;-\;\sqrt{A^{2}\;-\;G^{2}}}\end{array} \)

Since, x² – 2Ax + G² = 0,

Therefore, by using the quadratic formula, we get

\(\begin{array}{l}\mathbf{(a, b) = \frac{2\;A\;\pm \;\sqrt{4\;A^{2}\;-\;4\;G^{2}}}{2}}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{(a, b) = A\;\pm \;\sqrt{A^{2}\;-\;G^{2}}}\end{array} \)

📘 Applications of AM-GM Relationship

📌 Inequality Proofs in Algebra & Calculus
📌 Optimization Problems in Economics & Physics
📌 Financial Mathematics – Investment & Growth Analysis
📌 Data Science & Statistics – Predictive Modeling
📌 Engineering – Signal Processing & Circuit Analysis

Solved Examples

Example 1: The arithmetic mean of two positive numbers, a and b, is 1 more than its geometric mean. Find the value of a and b, if their difference is 8.

Solution:

According to the given condition,

a – b = 8

Or,

\(\begin{array}{l}\mathbf{\left ( \sqrt{a} \right )^{2}\;-\;\left ( \sqrt{b} \right )^{2}\;=\;8}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{\left ( \sqrt{a}\;-\;\sqrt{b} \right )\;\;\left ( \sqrt{a}\;+\;\sqrt{b} \right )\;=\;8}…..(1)\end{array} \)

And, AM – GM = 1 (given)

i.e.

\(\begin{array}{l}\mathbf{\frac{a\;+\;b}{2}\;-\;\sqrt{ab}\;=\;1}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{a\;+\;b\;-\;2\;\sqrt{ab}\;=\;2}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{\left ( \;\sqrt{a}\;-\;\sqrt{b}\; \right )^{2}\;=\;2}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{\;\sqrt{a}\;-\;\sqrt{b}\;=\;\pm \;2}….(2)\end{array} \)

Now, on substituting the values of equation (2) to equation (1), we get

\(\begin{array}{l}\mathbf{\left ( \pm\; 2 \right ) \;\;\left ( \sqrt{a}\;+\;\sqrt{b} \right )\;=\;8}\end{array} \)

\(\begin{array}{l}\mathbf{\sqrt{a}\;+\;\sqrt{b}\;=\;\pm \;4}….(3)\end{array} \)

Now, on squaring both the LHS and RHS of equations (2) and (3), we get

\(\begin{array}{l}\mathbf{a\;+\;b\;-\;2\;\sqrt{ab}\;=\;4}…..(4)\end{array} \)

And,

\(\begin{array}{l}\mathbf{a\;+\;b\;+\;2\;\sqrt{ab}\;=\;16}…..(5)\end{array} \)

On solving the equation (4) and equation (5), we get

ab = 9

so b = 9/a……(6)

And a + b = 10 . . . . . (7)

On substituting the values of b in equation (7), we get

a + 9/a = 10

\(\begin{array}{l}a^{2} – 10a + 9 = 0\end{array} \)

Now, using the quadratic formula

\(\begin{array}{l}\mathbf{a\;=\;\frac{10\;\pm \;\sqrt{100\;-\;36}}{2}}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{a\;=\;\frac{10\;\pm \;8}{2}\;=\;5\;\pm \;4}\end{array} \)

= 9 or 1

Hence, when a = 9, b = 1 and when a  = 1, b = 9

Therefore, the required numbers are 9 and 1.

Example 2: If the arithmetic and geometric mean of two positive real numbers a and b are 13 and 12, respectively. Find the value of a and b.

Solution:

Given: A = 13 and G = 12

Note: If A and G are arithmetic and geometric mean between two numbers a and b, then

\(\begin{array}{l}\mathbf{a\;=\;A\;+\;\sqrt{A^{2}\;-\;G^{2}}\;\;and\;\;b\;=\;A\;-\;\sqrt{A^{2}\;-\;G^{2}}}\end{array} \)

Therefore,

\(\begin{array}{l}\mathbf{a\;=\;13\;+\;\sqrt{13^{2}\;-\;12^{2}}\;\;and\;\;b\;=\;13\;-\;\sqrt{13^{2}\;-\;12^{2}}}\end{array} \)

Or,

\(\begin{array}{l}\mathbf{a\;=\;13\;+\;\sqrt{169\;-\;144}\;\;and\;\;b\;=\;13\;-\;\sqrt{169\;-\;144}}\end{array} \)

Therefore, a = 18, and b = 8

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