Problem 209 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) be in H.P. then \(a, b, c\) are also in H.P.

Step-by-Step Solution

Answer

Given that \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) are in H.P., we first find their reciprocals which are \(a(b+c), b(c+a), c(a+b)\). We check if these reciprocals are in A.P., and satisfy the equation \(2 \times b(c+a) = a(b+c) + c(a+b)\). Simplifying, we get \(\frac{1}{c} + \frac{2}{a} + \frac{1}{b} = \frac{2}{a} + \frac{2}{b} + \frac{2}{c}\). Rearranging and solving, we get the condition \(\frac{b-a}{ab} =\frac{c-b}{bc}\) which implies that a, b, and c are in H.P.

Step 1: Recalling the property of Harmonic Progression

If three numbers are in Harmonic Progression, then their reciprocals are in Arithmetic Progression. Given that \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) are in H.P., their reciprocals will be in A.P.

Step 2: Writing the reciprocals of the given expressions

The reciprocals of the given expressions are: \(a(b+c), b(c+a), c(a+b)\)

Step 3: Checking if the reciprocals are in Arithmetic Progression

Since the reciprocals of the given expressions are in A.P., the following condition needs to be satisfied: \[2 \times b(c+a) = a(b+c) + c(a+b)\]

Step 4: Simplifying the equation

Expanding the equation, we get: \[2bc + 2ab = ab + ac + cb + ac\] Simplifying further, we get: \[ab + 2bc + ac = 2(ac + bc + ab)\] Now, divide both sides of the equation by \(abc\): \[\frac{1}{c} + \frac{2}{a} + \frac{1}{b} = \frac{2}{a} + \frac{2}{b} + \frac{2}{c}\]

Step 5: Proving a, b, and c are in Harmonic Progression

The above equation implies the following: \[\frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c}\] Now, let's take the reciprocals of the terms involved: \[\frac{1}{(\frac{1}{a} - \frac{1}{b})} = \frac{1}{(\frac{1}{b}-\frac{1}{c})}\] Simplifying, we get: \[\frac{1}{\frac{b-a}{ab}} = \frac{1}{\frac{c-b}{bc}}\] So, we have: \[\frac{b-a}{ab} =\frac{c-b}{bc}\] Hence, observe that this is the condition for three numbers a, b, and c to be in H.P. Since we have proven the condition, we conclude that a, b, and c are in Harmonic Progression.

Key Concepts

Arithmetic Progression

Arithmetic progression, often abbreviated as A.P., is a sequence of numbers where the difference between any two consecutive terms is constant. This difference is known as the "common difference." For example, in the sequence 3, 6, 9, 12, the common difference is 3. Understanding arithmetic progression is crucial for tackling problems involving series and sequences.
To check if three numbers, say \( x, y, z \), are in A.P., the condition \( 2y = x + z \) should be satisfied. This expression ensures that the middle term is the average of the other two.
In our problem, the reciprocals of the terms given in harmonic progression can form an A.P., which helps us simplify complex algebraic expressions into recognizable patterns.

Reciprocals

A reciprocal of a number is its multiplicative inverse. For any non-zero number \( x \), its reciprocal is \( \frac{1}{x} \). When dealing with progressions, reciprocals play a particularly important role when transitioning between arithmetic progressions and harmonic progressions.
In a harmonic progression (H.P.), the sequence \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) helps us understand how the reciprocals of terms relate to each other. If these reciprocals are in A.P., we understand that the original terms are in H.P.
In our exercise, since \( \frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)} \) are in harmonic progression, their reciprocals \( a(b+c), b(c+a), c(a+b) \) form an arithmetic progression. This simple switch between reciprocals helps reveal patterns that are otherwise obscured.

Symmetric Functions

Symmetric functions are expressions that remain unchanged under any permutation of their variables. In mathematics, especially in algebra, these functions are useful in simplifying expressions and equations.
In the exercise problem, when you look at the reciprocals' arithmetic progression equation, we see an expression like \( a(b+c), b(c+a), c(a+b) \). These expressions are symmetric because swapping the variables among these will yield the same result.
By exploiting symmetry, we can solve the problem more efficiently, as symmetric functions often lead to simpler ways to verify conditions or prove relationships within algebraic structures.

Algebraic Manipulation

Algebraic manipulation is a collection of techniques used to transform complex equations into simpler or more useful forms. These techniques often involve expanding, factoring, or simplifying expressions using basic algebraic rules.
In the exercise, we applied algebraic manipulation when transforming the condition for the reciprocals to be in A.P., starting from \( 2b(c+a) = a(b+c) + c(a+b) \). We expanded this equation and rearranged terms to ultimately deduce that \( \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c} \).
Skilled algebraic manipulation allows us to reveal the underlying structure of a problem, making seemingly complex relationships between numbers more evident and easier to understand.