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Sequences and Series - Relationship Between A.M. and G.M.

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Definition of Arithmetic Mean (A.M.): For any two positive real numbers aa and bb, the Arithmetic Mean is the value A=a+b2A = \frac{a+b}{2}. Visually, on a number line, AA represents the exact midpoint between the points aa and bb.

Definition of Geometric Mean (G.M.): For any two positive real numbers aa and bb, the Geometric Mean is the value G=abG = \sqrt{ab}. Geometrically, if you have a rectangle with sides aa and bb, GG is the side of a square that has the same area as that rectangle.

The A.M.-G.M. Inequality: For any two positive real numbers aa and bb, the Arithmetic Mean is always greater than or equal to the Geometric Mean (AGA \ge G). On a coordinate plane, the value of AA will always be located at or to the right of GG on the horizontal axis.

Condition for Equality: The relationship A=GA = G occurs if and only if the two numbers are equal, i.e., a=ba = b. If aba \neq b, then AA is strictly greater than GG (A>GA > G).

Geometric Visualization via Semicircle: Consider a semicircle with a diameter of length a+ba+b. The radius of this circle is the A.M., which is a+b2\frac{a+b}{2}. If a perpendicular line is drawn from the point where segments aa and bb meet on the diameter to the circumference, the length of this perpendicular is the G.M., ab\sqrt{ab}. Since the radius is the maximum possible height within the semicircle, it visually demonstrates that A.M. \ge G.M.

Quadratic Equation Connection: If AA and GG are the A.M. and G.M. of two numbers aa and bb, then these numbers are the roots of the quadratic equation x22Ax+G2=0x^2 - 2Ax + G^2 = 0. This relates the means to the algebraic structure of polynomial roots.

Nature of Roots: Since AGA \ge G, the discriminant of the equation x22Ax+G2=0x^2 - 2Ax + G^2 = 0, which is D=4(A2G2)D = 4(A^2 - G^2), is always non-negative. This ensures that the numbers aa and bb are always real when AA and GG are given such that AGA \ge G.

📐Formulae

A=a+b2A = \frac{a+b}{2}

G=abG = \sqrt{ab}

AGA \ge G

AG=a+b2ab2=(ab)22A - G = \frac{a+b-2\sqrt{ab}}{2} = \frac{(\sqrt{a}-\sqrt{b})^2}{2}

x2(a+b)x+ab=0x^2 - (a+b)x + ab = 0

x22Ax+G2=0x^2 - 2Ax + G^2 = 0

Roots: x=A±A2G2x = A \pm \sqrt{A^2 - G^2}

💡Examples

Problem 1:

If the Arithmetic Mean (A.M.) of two positive numbers is 1010 and their Geometric Mean (G.M.) is 88, find the numbers.

Solution:

Step 1: Let the two numbers be aa and bb. We are given A=10A = 10 and G=8G = 8. Step 2: Use the quadratic equation formula for numbers when A.M. and G.M. are known: x22Ax+G2=0x^2 - 2Ax + G^2 = 0. Step 3: Substitute the values: x22(10)x+(8)2=0x220x+64=0x^2 - 2(10)x + (8)^2 = 0 \Rightarrow x^2 - 20x + 64 = 0. Step 4: Solve the quadratic equation by factoring: (x16)(x4)=0(x - 16)(x - 4) = 0. Step 5: The roots are x=16x = 16 and x=4x = 4. Therefore, the two numbers are 1616 and 44.

Explanation:

This approach uses the property that two numbers are the roots of a quadratic equation formed using their A.M. and G.M. Alternatively, one could solve the system a+b=20a+b=20 and ab=64ab=64.

Problem 2:

Prove that for any two positive real numbers aa and bb, the Arithmetic Mean is always greater than or equal to the Geometric Mean.

Solution:

Step 1: Consider the difference AG=a+b2abA - G = \frac{a+b}{2} - \sqrt{ab}. Step 2: Take a common denominator: AG=a+b2ab2A - G = \frac{a+b-2\sqrt{ab}}{2}. Step 3: Observe that the numerator is a perfect square: a+b2ab=(a)2+(b)22ab=(ab)2a+b-2\sqrt{ab} = (\sqrt{a})^2 + (\sqrt{b})^2 - 2\sqrt{a}\sqrt{b} = (\sqrt{a}-\sqrt{b})^2. Step 4: Thus, AG=(ab)22A - G = \frac{(\sqrt{a}-\sqrt{b})^2}{2}. Step 5: Since the square of any real number is non-negative, (ab)20(\sqrt{a}-\sqrt{b})^2 \ge 0. Therefore, AG0A - G \ge 0, which implies AGA \ge G.

Explanation:

This algebraic proof demonstrates that the difference between A.M. and G.M. is proportional to the square of the difference of the square roots of the numbers, which can never be negative.