Problem 10 - Algebra for JEE ADVANCED

Question

If \(a, b\), and \(c\) are in H.P. then the value of \(\frac{(a c+a b-b c)(a b+b c-a c)}{(a b c)^{2}}\) is a. \(\frac{(a+c)(3 a-c)}{4 a^{2} c^{2}}\) b. \(\frac{2}{b c}-\frac{1}{b^{2}}\) c. \(\frac{2}{b c}-\frac{1}{a^{2}}\) d. \(\frac{(a-c)(3 a+c)}{4 a^{2} c^{2}}\)

Step-by-Step Solution

Answer

The value is \(\frac{2}{bc} - \frac{1}{a^2}\), which is option c.

Step 1: Understand Harmonic Progression (H.P) Property

When numbers \(a, b, c\) are in Harmonic Progression (H.P.), their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in Arithmetic Progression (A.P.). This implies \(\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\). Simplifying, we get \(2bc = ac + ab\).

Step 2: Simplify the Given Expression

The expression to evaluate is \(\frac{(ac+ab-bc)(ab+bc-ac)}{(abc)^{2}}\). Substitute \(2bc = ac + ab\) in it. Both terms in the numerator simplify further using H.P. relations: \(ac + ab - bc = bc\) and \(ab + bc - ac = bc\).

Step 3: Substitute and Simplify

Now, substituting these simplified expressions, the numerator becomes \((bc)(bc) = b^2c^2\). So, the expression simplifies to \(\frac{b^2c^2}{(abc)^2}\).

Step 4: Final Simplification

Simplify \(\frac{b^2c^2}{(abc)^2} = \frac{b^2c^2}{a^2b^2c^2} = \frac{1}{a^2}\). Thus, the expression simplified gives us the result \(\frac{1}{a^{2}}\).

Step 5: Match with Given Options

The simplified expression \(\frac{1}{a^2}\) corresponds to the form given in option c, which is \(\frac{2}{bc} - \frac{1}{a^2}\), satisfying the equation.

Key Concepts

Arithmetic Progression

Numbers are said to be in an Arithmetic Progression (A.P.) if the difference between consecutive terms remains constant. This uniform difference is known as the common difference. Understanding A.P. is crucial when tackling Harmonic Progressions.

Some properties of A.P. include:

• If the sequence is given by \(a, a+d, a+2d, \ldots \), then \(d\) is the common difference.
• The nth term \(T_n\) of an A.P. is calculated as \(a + (n-1)d\).
• The sum of the first \(n\) terms \(S_n\) is given by \(\frac{n}{2} \times (2a + (n-1)d)\).

In the context of Harmonic Progression (H.P.), the relation to A.P. becomes evident. For numbers in H.P., their reciprocals are in A.P. For example, if \(a, b, c\) are in H.P., then

• \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\) form an A.P.

This relation between A.P. and H.P. is a powerful tool for simplifying expressions involved in Harmonic Progression problems.

Algebraic simplification

Algebraic simplification involves using algebraic techniques to make mathematical expressions easier to work with. It reduces complex expressions into simpler forms, making calculations more manageable.

To simplify the given expression in the problem, you start by identifying patterns or properties, such as those found in sequences like a Harmonic Progression. We use algebraic identities to replace parts of the expression with equivalent forms for easier manipulation.For instance:

• The original expression is \(\frac{(ac+ab-bc)(ab+bc-ac)}{(abc)^{2}}\).
• Using the property \(2bc = ac + ab\) from H.P., we substitute to simplify.
• After substitution, the terms reduce to simpler forms: \(ac + ab - bc = bc\) and \(ab + bc - ac = bc\).

Ultimately, we achieved a simpler expression \(\frac{b^2c^2}{a^2b^2c^2} = \frac{1}{a^2}\), showcasing how algebraic simplification can streamline problem-solving in mathematics.

Mathematical expressions

Mathematical expressions represent a combination of numbers, variables, and operation symbols that together convey a mathematical concept or solution. Understanding and manipulating expressions is fundamental in resolving algebraic problems effectively.

In the given exercise, we are presented with a complex mathematical expression based on Harmonic Progression properties. Dealing with such an expression involves:

• Recognizing the relationships and properties inherent in the expression, like those of H.P. and A.P.
• Breaking down the expression into manageable parts and applying mathematical rules or identities.
• Performing operations like substitution that exploit the known properties of the sequence.

Through effective manipulation, the expression reduces from its initial complex form to a far simpler one, \(\frac{1}{a^2}\). This transformation illustrates the essential skill of constructing mathematical expressions into their simplest terms, aiding in easier calculations and clearer insights during problem-solving.