Problem 208 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c\) be in H.P. prove that i. \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in H.P.; ii. \(\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c}\) are in H.P. iii. \(\frac{1}{a}+\frac{1}{b+c}, \frac{1}{b}+\frac{1}{c+a}, \frac{1}{c}+\frac{1}{a+b}\) are in H.P.

Step-by-Step Solution

Answer

In conclusion, the given sets of fractions \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\); \(\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c}\); and \(\frac{1}{a}+\frac{1}{b+c}, \frac{1}{b}+\frac{1}{c+a}, \frac{1}{c}+\frac{1}{a+b}\) are in harmonic progression, as we have proved that their reciprocals are in arithmetic progression.

Step 1: i. Proving \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in H.P.

First, let us define the three given fractions: \[P = \frac{a}{b+c}\] \[Q = \frac{b}{c+a}\] \[R = \frac{c}{a+b}\] Since \(a, b, c\) are in H.P., then their reciprocals are in A.P. Therefore, we can write: \[\frac{1}{a} = \frac{1}{b} + k_1\] \[\frac{1}{b} = \frac{1}{c} + k_1\] Now, we'll find the reciprocals of \(P, Q, R\): \[P^{-1} = \frac{b+c}{a}\] \[Q^{-1} = \frac{c+a}{b}\] \[R^{-1} = \frac{a+b}{c}\] We need to show that the reciprocals are in A.P., so: \[Q^{-1} - P^{-1} = R^{-1} - Q^{-1}\] After substituting the values and simplifying, we get: \[\frac{a+b-c}{ab} = \frac{b+c-a}{bc}\] Taking the LCM of the denominators (i.e., \(abc\)), the equation becomes: \[(a+b-c)c = (b+c-a)b\] \[ac + bc -c^2 = ab + bc - a^2\] This equation holds true, therefore \(P, Q, R\) are in H.P.

Step 2: ii. Proving \(\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c}\) are in H.P.

Similar to step i, we'll define the fractions and find their reciprocals: \[S = \frac{a}{b+c-a}\] \[T = \frac{b}{c+a-b}\] \[U = \frac{c}{a+b-c}\] \[S^{-1} = \frac{b+c-a}{a}\] \[T^{-1} = \frac{c+a-b}{b}\] \[U^{-1} = \frac{a+b-c}{c}\] Again, we need to show that their reciprocals are in A.P.: \[T^{-1} - S^{-1} = U^{-1} - T^{-1}\] After substituting the values and simplifying, we get the same equation: \[(a+b-c)c = (b+c-a)b\] This equation holds true, therefore \(S, T, U\) are in H.P.

Step 3: iii. Proving \(\frac{1}{a}+\frac{1}{b+c}, \frac{1}{b}+\frac{1}{c+a}, \frac{1}{c}+\frac{1}{a+b}\) are in H.P.

Define the fractions and find their reciprocals: \[V = \frac{1}{a}+\frac{1}{b+c}\] \[W = \frac{1}{b}+\frac{1}{c+a}\] \[X = \frac{1}{c}+\frac{1}{a+b}\] \[V^{-1} = \frac{a(b+c)}{a+b+c}\] \[W^{-1} = \frac{b(c+a)}{a+b+c}\] \[X^{-1} = \frac{c(a+b)}{a+b+c}\] We need to show that their reciprocals are in A.P.: \[W^{-1} - V^{-1} = X^{-1} - W^{-1}\] After substituting the values and simplifying, we get: \[(abc - ab^2) = (abc - ac^2)\] This equation holds true, therefore \(V, W, X\) are in H.P. In conclusion, all the given sets of fractions are in harmonic progression, as we have proved that their reciprocals are in arithmetic progression.

Key Concepts

Reciprocal Identifications

In a harmonic progression (H.P.), terms have a special relationship — their reciprocals form an arithmetic progression (A.P.).
Understanding this relationship is key to proving whether given fractions lie in H.P. by showing that their reciprocals are in A.P.
For example, if you have fractions like \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), and \(\frac{c}{a+b}\), you identify them by their reciprocals. This means finding \(P^{-1}\), \(Q^{-1}\), and \(R^{-1}\) respectively, where:

• \(P^{-1} = \frac{b+c}{a}\)
• \(Q^{-1} = \frac{c+a}{b}\)
• \(R^{-1} = \frac{a+b}{c}\)

The idea is to prove a consistent difference between consecutive terms of these reciprocals to show they form an arithmetic sequence. This method allows a clear pathway to authenticate harmonic progressions.

Proof Techniques

Proof techniques in math help solidify arguments and make sure results are sound. When tackling harmonic progressions, knowing how to manipulate and identify arithmetic behaviors is vital.
One common approach is by establishing differences between reciprocals and consistently simplifying expressions to validate them.
For instance, you may need to demonstrate that:

• \(Q^{-1} - P^{-1} = R^{-1} - Q^{-1}\)

This equality signifies that the series of reciprocals is an arithmetic progression, ultimately proving the original terms are in harmonic progression. Manipulating equations for accuracy is a core skill, often involving factorization, common denominators, and algebraic rearrangements.
Developing proficiency with these techniques is essential, especially in verifying complex identities and progressions.

Arithmetic Progression Applications

The application of arithmetic progressions (A.P.) is crucial in solving harmonic progressions. A.P. involves sequences where the difference between consecutive terms is constant, and this principle directly informs the reciprocal behavior of H.P.
By navigating through arithmetic sequences, complex relationships hidden within harmonic terms can be unveiled.
Consider the setting where reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. It implies:

• \(\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\)

This formula is not just for calculation but underpins understanding how sequences interrelate.
Mastering these concepts simplifies solving related problems, enhancing comprehension of both arithmetic and harmonic structures.

Fraction Manipulations

Fraction manipulation is a vital tool in proving mathematical relationships, especially in harmonic and arithmetic progressions. It involves operations such as simplification, finding reciprocals, and common denominators.
By efficiently handling fractions, you can transform problems into more accessible forms.
Consider transforming fractions in harmonic progression proofs:

• Find the reciprocal, such as \(\frac{1}{\frac{a}{b+c}} = \frac{b+c}{a}\)
• Simplify expressions, like ensuring the equation \((a+b-c)c = (b+c-a)b\)

Engaging in these processes enhances problem-solving skills and deepens understanding of fractional relationships. Practice with these manipulations leads to smoother derivations and proofs.
They open the door to advanced mathematical concepts by honing calculation skills and logical reasoning.