Problem 192 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A_{1}, A_{2}\) be two AM's and \(G_{1}, G_{2}\) be two GM's between \(a\) and \(b\), then prove that \(\frac{A_{1}+A_{2}}{G_{1} G_{2}}=\frac{a+b}{a b}\).

Step-by-Step Solution

Answer

We find expressions for arithmetic means \(A_1\) and \(A_2\) by evenly spacing them between 'a' and 'b': \[A_1 = a + \frac{b-a}{4}\] \[A_2 = b - \frac{b-a}{4}\]. Similarly, we find expressions for geometric means \(G_1\) and \(G_2\): \[G_1 = a \left(\frac{b}{a}\right)^{\frac{1}{4}}\] \[G_2 = b \left(\frac{a}{b}\right)^{\frac{1}{4}}\]. Substituting these expressions into the given equation and simplifying, we have \[\frac{A_1 + A_2}{G_1G_2} = \frac{a + b}{ab}\], which proves the result.

Step 1: Find expressions for A₁ and A₂

We are given two AM's, A₁ and A₂, between 'a' and 'b'. By definition, an AM between two numbers 'a' and 'b' can be defined as: \[\frac{a + b}{2}\] However, since we have two AMs, we need to find expressions for A₁ and A₂ using their properties. Since they are evenly spaced between 'a' and 'b', as per the definition of an arithmetic sequence, we can express A₁ and A₂ as such: \[A_1 = a + \frac{b-a}{4}\] \[A_2 = b - \frac{b-a}{4}\]

Step 2: Find expressions for G₁ and G₂

We are also given two GM's, G₁ and G₂, between 'a' and 'b'. By definition, a GM between two numbers 'a' and 'b' can be defined as: \[\sqrt{ab}\] Since there are two GMs, we need to find expressions for G₁ and G₂ using their properties. The geometric mean of two numbers is the square root of their product. For GM's in a sequence, we can express G₁ and G₂ as follows: \[G_1 = a \left(\frac{b}{a}\right)^{\frac{1}{4}}\] \[G_2 = b \left(\frac{a}{b}\right)^{\frac{1}{4}}\]

Step 3: Simplify the given equation using expressions for A₁, A₂, G₁, G₂

Now that we have expressions for A₁, A₂, G₁, and G₂, we can substitute these into the given equation and simplify: \[\frac{A_1 + A_2}{G_1G_2} = \frac{a + \frac{b-a}{4}+ b - \frac{b-a}{4}}{a \left(\frac{b}{a}\right)^{\frac{1}{4}} \cdot b \left(\frac{a}{b}\right)^{\frac{1}{4}}}\]

Step 4: Simplify and cancel terms

Simplify numerator and denominator: \[\frac{A_1 + A_2}{G_1G_2} = \frac{a + b}{ab}\]

Step 5: Compare the result with the RHS

By simplifying and canceling terms, we have obtained the equation: \[\frac{A_1 + A_2}{G_1G_2} = \frac{a + b}{ab}\] From this, we can see that the left-hand side (LHS) of the given equation is equal to the right-hand side (RHS). Therefore, the result is proven to be true.

Key Concepts

AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a fundamental concept in mathematics, especially in algebra and calculus. It states that for any non-negative numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). This inequality can be expressed as:

• For two numbers, say 'a' and 'b': \[ \frac{a + b}{2} \geq \sqrt{ab} \]
• For more numbers, the inequality extends similarly.

The AM-GM Inequality is particularly useful in optimization problems and proofs involving inequalities. Understanding its principle not only aids in handling numerical solutions but also provides insight into the behavior of sequences and series. In the problem above, using AM-GM principles helps to establish relationships between numbers (like the arithmetic means 'A' and geometric means 'G'). This concept is crucial in showing that mathematical relationships hold true across different expressions.

Sequences and Series

Sequences and Series are fundamental concepts in mathematics that deal with ordered lists of numbers and their sums, respectively.In sequences, numbers are arranged in a specific order. For example, in an arithmetic sequence, the difference between consecutive terms is constant, while in a geometric sequence, the ratio between consecutive terms is constant. Knowing these patterns allows us to derive specific formulas to calculate terms or sums of the series.

• An arithmetic sequence can be indicated by the n-th term formula: \[ a_n = a + (n-1)d \]where 'a' is the first term and 'd' is the common difference.
• A geometric sequence is expressed by: \[ a_n = a \cdot r^{n-1} \]where 'a' is the first term and 'r' is the common ratio.

These formulas are essential for solving problems involving series and understanding mathematical relationships, like those presented in the original exercise. Applying sequences and series principles helps us find intermediate points, such as arithmetic and geometric means.

Problem Solving in Calculus

Problem solving in calculus often involves understanding and manipulating various concepts like sequences, series, limits, derivatives, and integrals to find solutions. In calculus, integration is typically used to find areas under curves or solve problems involving accumulation, while differentiation focuses on rates of change. Key Problem Solving Strategies:

• Break down complex problems into simpler parts, which can be analyzed individually.
• Use known formulas and theorems, such as the AM-GM inequality or sequence formulas, to simplify expressions.
• Always check results by comparing different approaches to ensure validity and accuracy.

In the context of the original exercise, calculus problem-solving involves understanding and connecting various mathematical properties to prove the relationship between different mean values. By applying logical reasoning and algebraic manipulation, we successfully demonstrate the relationship between the AMs and GMs given in the problem.