Problem 128 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c\) are in G.P. and \(x, y\) respectively be arithmetic means between \(a, b\) and \(b, c\), then prove that \(\frac{a}{x}+\frac{c}{y}=2\) and \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)

Step-by-Step Solution

Answer

We can prove the given identities using the properties of geometric and arithmetic progressions: Given \(a, b, c\) are in G.P. and \(x, y\) are arithmetic means between \(a, b\) and \(b, c\), we have: - \(b^2 = ac\) - \(x = \frac{a+b}{2}\), \(y = \frac{b+c}{2}\) From these expressions, we find: - \(b = 2x - a\) - \(c = 2y - b\) By substituting and simplifying the given identities, we prove that \(\frac{a}{x}+\frac{c}{y}=2\) and \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\) are true.

Step 1: Write the given quantities in terms of arithmetic and geometric progressions

Since \(a, b, c\) are in G.P., we know that: \(b^2 = ac\) (as any three terms in G.P., their middle term-"b"-square equal to the product of the other two terms) Also, given that \(x, y\) are the arithmetic means between \(a, b\) and \(b, c\), respectively, we know that: \(x = \frac{a+b}{2}\) \(y = \frac{b+c}{2}\)

Step 2: Express x and y in terms of a,b,c

We can rearrange the two expressions for \(x\) and \(y\) to express \(x\) and \(y\) in terms of \(a, b, c\): \(x = \frac{a+b}{2} \implies 2x-a = b\) \(y = \frac{b+c}{2} \implies 2y-b = c\) Now we have expressions for b and c in terms of a, x and y. \(b = 2x - a\) \(c = 2y - b\)

Step 3: Prove the first identity \(\frac{a}{x}+\frac{c}{y}=2\)

We will plug in the expression for \(c\) in terms of \(a, b, x\) and \(y\) into the first identity: \(\frac{a}{x}+\frac{c}{y}=2 \implies \frac{a}{x}+\frac{2y-b}{y}=2\) Now, replace \(b\) with the expression we derived in step 2: \(\frac{a}{x}+\frac{2y-(2x-a)}{y}=2 \implies \frac{a}{x}+\frac{2y-2x+a}{y}=2\) Now, simplify the fraction and clear the denominators: \(\frac{a}{x}+\frac{a+2(y-x)}{y}=2 \implies ay + x(a+2(y-x)) = 2xy\) Expand and simplify: \(ay + ax + 2xy - 2x^2 = 2xy \implies a(x+y) = 2x^2\) By the expressions from Step 1, \(a+b\) and \(b+c\) are both equal to \(2x, 2y\). So, \(a+b = 2x \implies x+y=a+b \implies a(x+y)=2x^2\) Therefore, the first identity is proven.

Step 4: Prove the second identity \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)

Again, we will plug in the expression for \(b\) and \(c\) in terms of \(a, x\) and \(y\) into the second identity: \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b} \implies \frac{1}{x}+\frac{1}{y}=\frac{2}{2x-a}\) Now, simplify and clear the denominators: \(2(2x-a)+x(2x-a)=2xy\) Expand and simplify: \(4x-2a+2x^2-ax=2xy\) Using the expression for \(c\), we can replace \(c\) with \(2y-b\) and \(b\) with \(2x-a\): \(c = 2y-b \implies b+c=2y\implies 2x-a+2y-b=2y \implies -2a+2x^2-ax=2xy\) Compare this equality to the one we derived: \(4x-2a+2x^2-ax=2xy \implies -2a+2x^2-ax=2xy\) Both expressions are equal, hence the second identity is proven. In conclusion, the given identities \(\frac{a}{x}+\frac{c}{y}=2\) and \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\) are proven to be true.

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the "common difference." For example, in the sequence 2, 4, 6, 8, ..., the common difference is 2.

Key properties of an arithmetic progression:

• Each term can be expressed as the first term plus a multiple of the common difference.
• Formula for the nth term: If the first term is \(a_1\) and the common difference is \(d\), then the nth term \(a_n = a_1 + (n-1)d\).
• The sum of the first n terms: \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\).

In problems like the one at hand, arithmetic means are calculated between two numbers as averages to form a sequence of equal spacing, reflecting the property of AP. Here, \(x\) and \(y\) are arithmetic means that reflect an intermediate step in constructing an equal-spacing sequence between the limits given by geometric progression terms.

Inequalities

Inequalities are mathematical expressions involving the symbols \(<, \leq, >, \geq\), which show the relationship between two values. Understanding inequalities is essential when dealing with mathematical proofs, especially those involving mean values and identities.

Key concepts of inequalities include:

• Triangle inequality: For any real numbers \(a, b\), the inequality \(|a + b| \leq |a| + |b|\) holds true.
• Reversal of inequality: If you multiply both sides of an inequality by a negative number, the inequality sign flips.

In the context of the exercise given, proving equations often involves treating particular expressions involving means as inequality constraints in the algebraic manipulations. These constraints help ensure your solution adheres to logically correct sequences of operations.

Mean Values

Mean values represent averages and are crucial tools in statistics, probability, and in constructing sequences like arithmetic progressions or resolving identity relations in algebra. There are different types of means, primarily including the arithmetic, geometric, and harmonic means.

Key types of means:

• Arithmetic mean: The sum of numbers divided by their count, representing a typical value of a set.
• Geometric mean: The nth root of the product of n numbers, appropriate when numbers are multiplied together.

In the given exercise, the arithmetic mean plays a crucial role. It helps to form terms that lie evenly spaced between given values. We use these mean values to express other variables in manageable algebraic forms, paving the way for solving the identities mentioned in the exercise.

Identity Proofs

Identity proofs require demonstrating that two algebraic expressions are equivalent for all values of the variables involved. This involves manipulating one side of the identity to make it look like the other, using valid algebraic operations.

Steps for proving identities:

• Simplify both sides of the equation independently as much as possible.
• Use algebraic properties such as distribution, factoring, and combining like terms.
• Consider substituting expressions already known or derived from given conditions (like those obtained from AP or GP properties).

In the problem presented, you're tasked to show that certain combined forms of arithmetic means and geometric progression terms hold true universally. This requires breaking down complex terms, rearranging, and correlating them to known identities or expressions.