Relation Between AM and GM and Related Problems – Class 11 Applied Mathematics

Arithmetic Mean (AM) and Geometric Mean (GM)

For any two positive real numbers a and b, the Arithmetic Mean is defined as (a + b)/2 and the Geometric Mean is defined as √ab. These measures provide central tendencies for data sets. The relationship between them is fundamental in establishing inequalities in algebra.

AM = (a + b)/2, GM = √ab

Example: Solved Example: Find AM and GM of 9 and 4

Show Step-by-Step Solution

Step 1: AM = (9 + 4)/2 = 13/2 = 6.5
Step 2: GM = √(9 × 4) = √36 = 6
Answer: AM = 6.5, GM = 6

The AM ≥ GM Inequality

For any two positive real numbers a and b, the Arithmetic Mean is always greater than or equal to the Geometric Mean. The equality holds if and only if a = b. This property is widely used to find the minimum values of expressions.

(a + b)/2 ≥ √ab

Example: Solved Example: Verify AM ≥ GM for 8 and 2

Show Step-by-Step Solution

Step 1: AM = (8 + 2)/2 = 5
Step 2: GM = √(8 × 2) = √16 = 4
Step 3: Since 5 ≥ 4, the inequality holds.
Answer: Verified

Finding Numbers Given AM and GM

If the AM and GM of two numbers are known, the numbers can be found by solving a quadratic equation. If AM = A and GM = G, the numbers are the roots of the equation x² - 2Ax + G² = 0.

x² - (2A)x + G² = 0

Example: Solved Example: Find numbers if AM = 5 and GM = 4

Show Step-by-Step Solution

Step 1: Form equation x² - 2(5)x + 4² = 0
Step 2: x² - 10x + 16 = 0
Step 3: (x - 8)(x - 2) = 0
Answer: The numbers are 8 and 2

Section A: Multiple Choice Questions (MCQs)

easyIf a = 16 and b = 4, the AM is:

A. 8 B. 10 C. 12 D. 64

Show Answer

✔ Correct: B

AM = (16 + 4)/2 = 20/2 = 10.

easyIf a = 9 and b = 1, the GM is:

A. 3 B. 5 C. 9 D. 10

Show Answer

✔ Correct: A

GM = √(9 × 1) = √9 = 3.

mediumIf AM = 10 and GM = 8, the quadratic equation is:

A. x² - 10x + 64 = 0 B. x² - 20x + 64 = 0 C. x² - 20x + 8 = 0 D. x² - 10x + 8 = 0

Show Answer

✔ Correct: B

Equation is x² - 2Ax + G² = 0. Here 2A = 20 and G² = 64.

mediumFor a = 2 and b = 8, the difference AM - GM is:

A. 0 B. 1 C. 2 D. 5

Show Answer

✔ Correct: B

AM = 5, GM = 4. AM - GM = 5 - 4 = 1.

hardIf the AM of two numbers is 13 and their GM is 12, the numbers are:

A. 10, 16 B. 9, 17 C. 4, 22 D. 9, 16

Show Answer

✔ Correct: D

x² - 26x + 144 = 0. Roots are 9 and 16.

hardThe minimum value of x + 1/x for x > 0 is:

A. 1 B. 2 C. 0 D. 4

Show Answer

✔ Correct: B

Using AM ≥ GM: (x + 1/x)/2 ≥ √(x × 1/x) = 1. So x + 1/x ≥ 2.

very hardIf AM of two numbers is 5 and their GM is 3, the difference between the numbers is:

A. 4 B. 6 C. 8 D. 10

Show Answer

✔ Correct: C

x² - 10x + 9 = 0. Roots are 9 and 1. Difference = 9 - 1 = 8.

Section B: Assertion–Reason Questions

Assertion (A): For numbers 4 and 9, AM is 6.5 and GM is 6.
Reason (R): AM is always greater than or equal to GM for positive numbers.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Both statements are true and the reason correctly explains the relationship.

Assertion (A): The minimum value of 4x + 9/x for x > 0 is 12.
Reason (R): By AM-GM inequality, (4x + 9/x)/2 ≥ √(4x × 9/x) = √36 = 6.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Since (4x + 9/x)/2 ≥ 6, then 4x + 9/x ≥ 12.

Section C: Short Answer Questions

mediumGiven two numbers a = 25 and b = 9, find: (i) AM (ii) GM (iii) AM - GM

Show Answer & Solution

(i) AM = (25 + 9)/2 = 17
(ii) GM = √(25 × 9) = √225 = 15
(iii) AM - GM = 17 - 15 = 2

mediumGiven AM = 7.5 and GM = 6, find: (i) Sum of numbers (ii) Product of numbers (iii) The two numbers

Show Answer & Solution

(i) Sum = 2 × AM = 15
(ii) Product = GM² = 36
(iii) x² - 15x + 36 = 0 gives (x-12)(x-3)=0. Numbers are 12 and 3.

hardFor x > 0, consider the expression 9x + 1/x. Find: (i) AM of terms (ii) GM of terms (iii) Minimum value of expression

Show Answer & Solution

(i) AM = (9x + 1/x)/2
(ii) GM = √(9x × 1/x) = √9 = 3
(iii) Since AM ≥ GM, (9x + 1/x)/2 ≥ 3, so 9x + 1/x ≥ 6.

Section D: Long Answer Questions

hardIf the AM of two positive numbers is 10 and their GM is 8, find the numbers and verify the AM ≥ GM inequality.

Show Answer & Solution

Step 1: Let numbers be a and b. AM = (a+b)/2 = 10, so a+b = 20.
Step 2: GM = √ab = 8, so ab = 64.
Step 3: Form quadratic: x² - 20x + 64 = 0.
Step 4: Factorize: (x-16)(x-4) = 0. Numbers are 16 and 4.
Step 5: Verify: AM = (16+4)/2 = 10; GM = √64 = 8. Since 10 ≥ 8, inequality holds.

very hardProve that for any positive real number x, the expression (x + 2)(x + 8) has a minimum value when x is positive. Find that minimum value.

Show Answer & Solution

Step 1: Expand the expression: x² + 10x + 16.
Step 2: To find minimum, use AM-GM on x and 16/x terms if possible, or complete the square.
Step 3: x² + 10x + 16 = (x+5)² - 9.
Step 4: Since (x+5)² ≥ 0, the minimum value is -9 when x = -5 (but x > 0).
Step 5: Re-evaluating: For x > 0, the expression increases as x increases. The infimum as x → 0 is 16.

Section E: Case-Based Question

★

Case Study: A farmer has two rectangular plots. Plot A has dimensions 16m by 4m. Plot B has dimensions 10m by 10m. He wants to compare the average side lengths and the geometric mean of the sides.

Sub-question 1: (i) Find the AM and GM of the sides of Plot A. (ii) Find the AM and GM of the sides of Plot B.

Show Solution

(i) AM = (16+4)/2 = 10, GM = √64 = 8.
(ii) AM = (10+10)/2 = 10, GM = √100 = 10.

Sub-question 2: Compare the results. What do you observe about the AM and GM when the sides are equal?

Show Solution

In Plot A, AM > GM (10 > 8). In Plot B, AM = GM (10 = 10). Observation: AM = GM if and only if the numbers are equal.

CBSE Board & Sample Paper Past Year Questions

📋 CBSE 2023 📝 1 Mark

If AM of two numbers is 5 and GM is 4, the numbers are:

Show Answer & Solution

x² - 10x + 16 = 0. Roots are 8 and 2.

📋 CBSE 2022 📝 1 Mark

State the condition under which AM = GM for two positive numbers a and b.

Show Answer & Solution

The condition is a = b.

📋 CBSE Sample Paper 2024 📝 2 Marks

Find the minimum value of x + 25/x for x > 0.

Show Answer & Solution

(x + 25/x)/2 ≥ √25 = 5. So x + 25/x ≥ 10.

📋 CBSE 2020 📝 4 Marks

If the AM of two numbers is 10 and GM is 6, find the numbers.

Show Answer & Solution

Sum = 20, Product = 36. x² - 20x + 36 = 0. Roots are 10 ± √64 = 10 ± 8. Numbers are 18 and 2.

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