Question

If \(a, b, c\) are in H.P. then the value of \(\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{a}-\frac{1}{b}\right)\) is : (a) \(\frac{3}{b^{2}}-\frac{2}{a b}\) (b) \(a^{2}+b^{2}+c^{2}\) (c) \(\frac{2}{b c}-\frac{1}{b^{2}}\). (d) both (a) and (c)

Short answer

Given that a, b, and c are in harmonic progression, find the value of the expression: $$\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{a}-\frac{1}{b}\right)$$ The simplification of the given expression results in the following expression: $$\frac{4}{a}-\frac{4}{b}+\frac{2}{a^2}-\frac{4}{ab}+\frac{2}{b^2}$$ This expression matches option (d), which is a combination of options (a) and (c): $$\left(\frac{3}{b^{2}}-\frac{2}{a b}\right) + \left(\frac{2}{b c}-\frac{1}{b^{2}}\right)$$ Therefore, the correct answer is (d) both (a) and (c).

Step-by-step solution

Write the arithmetic progression terms of reciprocals

Since \(a\), \(b\), and \(c\) are in harmonic progression, their reciprocals are in arithmetic progression. We can write the reciprocals: $$\frac{1}{a} = x - d, \frac{1}{b} = x, \frac{1}{c} = x + d$$ where \(x\) is the middle term and \(d\) is the common difference.

Calculate the given expression using the arithmetic progression terms

Substitute the reciprocals in the given expression: $$\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{a}-\frac{1}{b}\right) = \left(x+(x+d)-(x-d)\right)\left((x+d)+(x-d)-x\right)$$

Simplify the expression

Let's simplify the expression obtained in step 2 by evaluating the terms inside the parentheses: $$\left(2x+d\right)\left(2d\right) = 4xd + 2d^2$$

Substitute \(x\) using terms of the original progression

Recall that \(\frac{1}{b} = x\). Substitute this back into our expression and simplify: $$4\cdot \frac{1}{b}d + 2d^2 = \frac{4d}{b} + 2d^2$$

Find the common difference in terms of \(a\) and \(b\)

Using the expressions for the reciprocals of \(a\) and \(b\), we find the common difference: $$\frac{1}{a} = x - d \implies x = d + \frac{1}{a}$$ Substitute this value of \(x\) into the expression for \(\frac{1}{b}\): $$\frac{1}{b} = d + \frac{1}{a} \implies d = \frac{1}{b} - \frac{1}{a}$$

Substitute the common difference expression into the original expression

Now, substitute the expression for \(d\) found in step 5 into the simplified expression from step 4: $$\frac{4\left(\frac{1}{b} - \frac{1}{a}\right)}{b} + 2\left(\frac{1}{b} - \frac{1}{a}\right)^2$$

Calculate and compare with the given options

Evaluate the expression and compare it with the given options: $$\frac{4}{a}-\frac{4}{b}+\frac{2}{a^2}-\frac{4}{ab}+\frac{2}{b^2}$$ This expression matches option (d), which is a combination of options (a) and (c): $$\left(\frac{3}{b^{2}}-\frac{2}{a b}\right) + \left(\frac{2}{b c}-\frac{1}{b^{2}}\right)$$ Therefore, the correct answer is (d) both (a) and (c).

Key concepts

Arithmetic Progression

Understanding arithmetic progression is paramount when dealing with series and sequences in mathematics. An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term. To clarify with an example: if we have an AP with a first term of 5 and a common difference of 3, the sequence would be 5, 8, 11, 14, and so forth.

In the exercise provided, the concept of AP is used inversely through the reciprocals of a harmonic progression. This application helps us simplify complex expressions by providing a structure to work with. The identification of the common difference and the ability to manipulate terms within an AP are crucial skills in solving such exercises.

Sequence and Series

The terms sequence and series are often encountered in mathematical problems dealing with ordered lists of numbers and their summations, respectively. A sequence is simply an ordered list of numbers following a particular rule, while a series is the sum of a sequence of numbers.

Our exercise involves understanding a type of sequence known as a harmonic progression (HP), which essentially means the reciprocals of the sequence's terms form an arithmetic progression. Consequently, when dealing with problems involving sequences and series, it's essential to identify the underlying pattern or rule governing the numbers, as it aids in making calculations and predictions about the sequence's behavior. For instance, an understanding of HP and AP in this case allows for the transformation and simplification of a series into a more manageable form.

Quantitative Aptitude

The term quantitative aptitude generally signifies one's ability to handle numerical and mathematical computations effectively. It is a critical skill for a wide range of academic disciplines and competitive exams, including those mentioned in our exercise, like solving for sequences in HP or AP.

Developing quantitative aptitude involves practicing various types of problems and understanding multiple concepts and techniques. Good quantitative aptitude enables students to approach algebraic expressions and complex equations with confidence. In the context of our problem, it is the quantitative aptitude of the student that permits efficient manipulation of algebraic expressions and the ability to solve for unknowns within those sequences, leading to the correct answer.