Problem 12 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c\) are in A.P., prove that the following are also in A.P.:- i. \(\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}\). ii. \(b+c, c+a, a+b\). iii. \(a^{2}(b+c), b^{2}(c+a), c^{2}(a+b)\). iv. \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), \quad c\left(\frac{1}{a}+\frac{1}{b}\right)\). v. \(\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}\).

Step-by-Step Solution

Answer

In each case, we found the differences between the second and the first number, and between the third and the second number, and after simplifying, we checked if they were equal. As a result, we proved that all the given expressions, i.e., \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\), \(b+c, c+a, a+b\), \(a^{2}(b+c), b^{2}(c+a), c^{2}(a+b)\), \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\), and \(\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}\), satisfy the conditions for being in Arithmetic Progression.

Step 1: Find the differences

Find the difference between the second and first numbers, and also between the third and second numbers: \[\frac{1}{ca} - \frac{1}{bc} \quad \text{and} \quad \frac{1}{ab} - \frac{1}{ca}.\]

Step 2: Take LCM and simplify

To subtract the fractions, take the LCM of denominators (in this case, it will be \(abc\)): \[\frac{b - a}{abc} \quad \text{and} \quad \frac{c - b}{abc}.\]

Step 3: Check for equal differences

Observe that the differences are equal since \(a, b, c\) are in A.P., which implies \((c - b) = (b - a)\). Therefore, \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P. ii. Prove \(b+c, c+a, a+b\) are in A.P.

Step 4: Find the differences

Find the difference between the second and first numbers, and also between the third and second numbers: \[(c+a) - (b+c) \quad \text{and} \quad (a+b) - (c+a).\]

Step 5: Simplify

Simplify the differences: \[a - b \quad \text{and} \quad b - c.\]

Step 6: Check for equal differences

Notice that the differences are equal since \(a, b, c\) are in A.P., which implies \(a - b = b - c\). Therefore, \(b+c, c+a, a+b\) are also in A.P. iii. Prove \(a^{2}(b+c), b^{2}(c+a), c^{2}(a+b)\) are in A.P.

Step 7: Find the differences

Find the difference between the second and first numbers, and also between the third and second numbers: \[b^2(c+a) - a^2(b+c) \quad \text{and} \quad c^2(a+b) - b^2(c+a).\]

Step 8: Factor and simplify

Factor and simplify the differences: \[ab(b-a) \quad \text{and} \quad ac(c-b).\]

Step 9: Check for equal differences

Notice that the differences are equal since \(a, b, c\) are in A.P., which implies \(a - b = c - b\). Therefore, \(a^{2}(b+c), b^{2}(c+a), c^{2}(a+b)\) are also in A.P. iv. Prove \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.

Step 10: Find the differences

Find the difference between the second and first numbers, and also between the third and second numbers: \[b\left(\frac{1}{c}+\frac{1}{a}\right) - a\left(\frac{1}{b}+\frac{1}{c}\right) \quad \text{and} \quad c\left(\frac{1}{a}+\frac{1}{b}\right) - b\left(\frac{1}{c}+\frac{1}{a}\right).\]

Step 11: Multiply and simplify

Multiply and simplify the differences: \[ac(a-b) \quad \text{and} \quad ab(b-c).\]

Step 12: Check for equal differences

Notice that the differences are equal since \(a, b, c\) are in A.P., which implies \(a - b = b - c\). Therefore, \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are also in A.P. v. Prove \(\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}\) are in A.P.

Step 13: Rationalize denominators

Rationalize the denominators using the conjugate: \[\frac{\sqrt{c}-\sqrt{b}}{a}, \frac{\sqrt{a}-\sqrt{c}}{b}, \quad \text{and} \quad \frac{\sqrt{b}-\sqrt{a}}{c}.\]

Step 14: Find the differences

Find the difference between the second and first numbers, and also between the third and second numbers: \[\frac{\sqrt{a}-\sqrt{c}}{b} - \frac{\sqrt{c}-\sqrt{b}}{a} \quad \text{and} \quad \frac{\sqrt{b}-\sqrt{a}}{c} - \frac{\sqrt{a}-\sqrt{c}}{b}.\]

Step 15: Take LCM and simplify

Take the LCM of the denominators and simplify differences: \[\frac{a \sqrt{a} - b \sqrt{b} - c \sqrt{c} + a \sqrt{c}}{ab} \quad \text{and} \quad \frac{a \sqrt{a} - b \sqrt{b} - c \sqrt{c} + b \sqrt{c}}{bc}.\]

Step 16: Check for equal differences

Observe that the differences are equal since \(a, b, c\) are in A.P., which implies \(a \sqrt{a} - b \sqrt{b} - c \sqrt{c}\) is a common factor. Therefore, \(\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}\) are also in A.P.

Key Concepts

Sequences and Series

Sequences and series are fundamental concepts in mathematics often used to describe a list of numbers following a particular pattern. Arithmetic Progression (A.P.) is a type of sequence where the difference between each consecutive term remains constant. This constant difference is referred to as the 'common difference', denoted by 'd'.
For example, the sequence 2, 4, 6, 8 is an arithmetic progression with a common difference of 2. Understanding sequences and series helps in solving complex problems and identifying patterns in data.
In your exercise, terms such as \(a, b, c\) are said to be in A.P. This means that the difference \(b-a = c-b\). Using this property, you can transform, manipulate or extend the sequence, as seen in the provided step-by-step solution.

Rational Functions

Rational functions are expressions that stand for the ratio of two polynomials. These functions are very important in algebra and calculus. A rational function is typically written as \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\) is not zero.
In the exercise, terms like \(\frac{1}{bc}\) and \(\frac{1}{ca}\) represent rational expressions. To prove these terms are in A.P., you rationalize the fractions and simplify by finding a common denominator. This helps reveal if the differences between terms in the sequence satisfy the condition for an arithmetic progression.

Difference of Squares

The difference of squares is a specific algebraic identity, expressed as \(a^2 - b^2 = (a-b)(a+b)\). This identity is a valuable tool in factorization and simplification of expressions. It allows us to break down complex expressions into simpler, more manageable components.
In solving parts of your exercise, you can identify patterns that follow this identity to simplify the problems. Recognizing when numbers or expressions fit this pattern helps navigate through seemingly complicated steps, bringing clarity to checking if formed differences meet the arithmetic progression criteria.

Coefficients in Polynomial Expressions

Coefficients are numbers placed in front of variables in polynomials and other algebraic expressions. They play a significant role in the solutions to polynomial equations, impacting the expression's value depending on the variable's power.
In your problem, expressions like \(a^2(b+c)\) and \(b^2(c+a)\) have coefficients in the form of squared terms (\(a^2\), \(b^2\), etc.). Identifying coefficients allows us to factor and simplify expressions, as demonstrated through factoring steps in the exercise. It's crucial in proving a sequence is an A.P., because the constant differences rely on such simplifications.