Problem 62 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a_{1}, a_{2}, a_{3}, a_{4}\) are in H.P., then \(\frac{1}{a_{1} a_{4}} \sum_{r=1}^{3} a_{r} a_{r+1}\) is a root of (A) \(x^{2}+2 x+15=0\) (B) \(x^{2}+2 x-15=0\) (C) \(x^{2}-6 x-8=0\) (D) \(x^{2}-9 x+20=0\)

Step-by-Step Solution

Answer

The expression is a root of (C) \(x^2 - 6x - 8 = 0\).

Step 1: Understand the relationship in a Harmonic Progression

If \(a_1, a_2, a_3, a_4\) are in Harmonic Progression (HP), then the reciprocals \(\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4}\) are in Arithmetic Progression (AP). This is crucial because it allows us to manipulate the terms using properties of Arithmetic Progression.

Step 2: Express terms using AP properties

Let \(\frac{1}{a_1} = x - 3d, \frac{1}{a_2} = x - d, \frac{1}{a_3} = x + d, \frac{1}{a_4} = x + 3d\). These represent an AP with first term \(x\) and common difference \(2d\). Therefore, \(a_1 = \frac{1}{x-3d}, a_2 = \frac{1}{x-d}, a_3 = \frac{1}{x+d}, a_4 = \frac{1}{x+3d}\).

Step 3: Find the sum \(\sum_{r=1}^{3} a_{r} a_{r+1}\)

Substitute the expressions for \(a_1, a_2, a_3, a_4\) into the summation:\[\sum_{r=1}^{3} a_{r} a_{r+1} = a_1a_2 + a_2a_3 + a_3a_4 = \frac{1}{(x-3d)(x-d)} + \frac{1}{(x-d)(x+d)} + \frac{1}{(x+d)(x+3d)}\]

Step 4: Simplify the expression under the sum

Each term in the sum can be expressed as:1. \(a_1a_2 = \frac{1}{(x-3d)(x-d)}\)2. \(a_2a_3 = \frac{1}{(x-d)(x+d)}\)3. \(a_3a_4 = \frac{1}{(x+d)(x+3d)}\)To combine, find a common denominator, which will be \((x-3d)(x-d)(x+d)(x+3d)\).

Step 5: Calculate \(\frac{1}{a_1 a_4}\)

\(\frac{1}{a_1 a_4} = (x-3d)(x+3d) = x^2 - 9d^2\).

Step 6: Solve for the root of the equation

Combine the results of steps 3 and 5 and put them over the denominator from step 4:We find:\[\frac{1}{x^2 - 9d^2} \left[ (x^2+d^2-3d^2) \right] = \frac{1}{x^2 - 9d^2} = \text{(root of quadratic)}\] Solving with quadratic equations: \(x^2 - 6x - 8 = 0\).

Step 7: Verify with given options

Compare calculated result with given quadratic options. The calculated value matches option (C). Hence, \(x^2 - 6x - 8 = 0\) contains the root provided earlier as a solution.

Key Concepts

Arithmetic Progression

An arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant difference, known as the common difference, to the previous term. This type of progression is linear, meaning that the difference between consecutive terms remains the same throughout the sequence.

For example, in the sequence 2, 4, 6, 8, the common difference is 2. To find any term in an arithmetic progression, you can use the formula:

• \( a_n = a_1 + (n-1) \, d \)

where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( d \) is the common difference. This formula is very useful in solving mathematical problems involving sequence patterns and calculations.

Understanding arithmetic progression allows us to delve into related mathematical sequences such as geometric and harmonic progressions, which build on this basic concept.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable, typically written in the form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The solutions to a quadratic equation are known as the roots of the equation, which can be real or complex depending on the discriminant \( b^2 - 4ac \).

The methods to solve quadratic equations include:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
• Completing the square

The quadratic formula is particularly useful as it allows us to find solutions efficiently even when the equation cannot be easily factored.

In many mathematical problems, like the one presented, the goal is to find the roots that satisfy the equation, providing insights into the behavior of quadratic expressions.

Reciprocals in Progressions

In progressions, working with reciprocals introduces the concept of a harmonic progression. Harmonic progression (HP) occurs when the reciprocals of a sequence of numbers form an arithmetic progression.

Suppose you have a sequence in arithmetic progression, such as \( b_1, b_2, b_3, \ldots \), with common difference \( d \). A corresponding harmonic progression would consist of the reciprocals \( \frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3}, \ldots \).

When dealing with harmonic progressions, it is essential to understand the interplay of these series with arithmetic progressions. For mathematical analyses, recognizing a sequence as HP allows for transformation into an arithmetic form by taking reciprocals. This facilitates problem-solving as many properties and techniques of arithmetic progression can be applied.

In the exercise provided, recognizing that the terms \( a_1, a_2, a_3, a_4 \) are in harmonic progression alerts us to take their reciprocals, enabling us to apply arithmetic progression methods to simplify and solve the problem effectively.