Problem 210 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(b+c, c+a, a+b\) are in H.P. then prove that i. \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in A.P.; ii. \(a^{2}, b^{2}, c^{2}\) are in A.P.

Step-by-Step Solution

Answer

In summary, we have proven that: i. \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in A.P., by expressing the reciprocals of these fractions and using the relationship between H.P. and A.P. terms, and ii. \(a^2, b^2, c^2\) are in A.P., by using algebraic manipulation and the relationship between A.P. and H.P. terms as well.

Step 1: Relationship between H.P. and A.P. terms

If \(x, y, z\) are in H.P., their reciprocals \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) will be in A.P. Conversely, if \(p, q, r\) are in A.P., then their reciprocals \(\frac{1}{p}, \frac{1}{q}, \frac{1}{r}\) will be in H.P.

Step 2: Writing the given terms as reciprocals of an A.P.

Since it is given that \(b+c, c+a, a+b\) are in H.P., their reciprocals will be in A.P. Thus, we have: \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.

Step 3: Proving that the given fractions are in A.P.

We'll write the three given fractions and find their reciprocals: \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) Taking reciprocals, we get: \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) We already have step 2 which states that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P. Thus, \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are indeed in A.P., and the first part (i) is proven.

Step 4: Algebraic manipulation to obtain the squares

We'll use the relationship between A.P. and H.P. terms once again to prove (ii). \((a+b)(b+c), (b+c)(c+a), (c+a)(a+b)\) are in A.P. We need the terms to be squares, so let's start by expanding the above expressions: \(ab+2bc+ac, bc+2ac+ab, ac+2ab+bc\) To make them into the squares of \(a, b, c\), we'll subtract a common term \(2abc\) from each expression: \(ab+2bc+ac-2abc, bc+2ac+ab-2abc, ac+2ab+bc-2abc\) Now the expressions become: \(a(a+b+c-2bc), b(a+b+c-2ac), c(a+b+c-2ab)\) We can observe that: \(a(a+b+c-2bc) = a^{2}(1-\frac{2bc}{a+b+c})\) \(b(a+b+c-2ac) = b^{2}(1-\frac{2ac}{a+b+c})\) \(c(a+b+c-2ab) = c^{2}(1-\frac{2ab}{a+b+c})\) Since \(1-\frac{2bc}{a+b+c}, 1-\frac{2ac}{a+b+c}, 1-\frac{2ab}{a+b+c}\) are positive constants, the expressions are now proportional to the squares of \(a, b, c\). As a result, we can conclude that \(a^2, b^2, c^2\) are in A.P., and the second part (ii) is proven.

Key Concepts

Arithmetic Progression

An Arithmetic Progression (A.P.) is a sequence of numbers where there is a constant difference between consecutive terms. This difference is often denoted as "d". For instance, in the sequence 2, 4, 6, 8, each term increases by 2. This simple structure makes it easy to calculate any term in the series using the expression:

• The nth term: \(a_n = a_1 + (n-1)d\)

An A.P. is central in solving problems involving sequences, as it provides a systematic approach. In many problems, like this one, recognizing the pattern of an A.P. helps transform or simplify sequences to reach conclusions.

Reciprocals in Progressions

Understanding the relationship between different types of progressions is crucial. In particular, when dealing with Harmonic Progression (H.P.) and Arithmetic Progression (A.P.), their reciprocals reveal much information.
If three terms \(x, y, z\) are in Harmonic Progression, their reciprocals \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) will form an Arithmetic Progression.

• Conversely, if three terms \(p, q, r\) are in A.P., their reciprocals \(\frac{1}{p}, \frac{1}{q}, \frac{1}{r}\) will be in H.P.

This understanding simplifies the approach to solving problems and allows easy manipulation of terms into different progressions.

Algebraic Manipulation

Algebraic manipulation involves rearranging and adjusting equations or expressions to find desired patterns or solutions. In solving problems related to progressions, this skill is vital.
Consider the task of proving sequences in H.P. or A.P., where expressions may need to be expanded or simplified to reveal hidden patterns.

• For example, manipulating expressions like \((a+b)(b+c), (b+c)(c+a), (c+a)(a+b)\) helps identify whether they belong to a particular progression.

The key is recognizing equalities or constants that may simplify whole expressions towards a known structure, such as an Arithmetic Progression.

Proving Series in Progressions

Proving that a series belongs to a specific progression often involves showing consistent differences (or ratios) between terms. For Arithmetic Progressions, it means proving

• The successive differences are equal.

In many problems, this involves algebraic steps to transform the original terms into forms that make the constant differences evident. This can include distributing terms, factoring, or using identities.
When it's about proving that specific powered terms (like squares) form A.P., algebraic manipulation is essential to demonstrate that transformations hold through all operations.
In essence, the proof rests on the clarity and consistency of patterns among terms, indicating their fit within the structural rules of the progression.