Problem 93 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a_{m}\) be the \(m\) th term of an A.P., then \(a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+\ldots .+a_{2 n-1}^{2}-a_{2 n}^{2}=\) (A) \(\frac{n-1}{2 n-1}\left(a_{1}^{2}-a_{2 n}^{2}\right)\) (B) \(\frac{n}{2 n-1}\left(a_{2 n}^{2}-a_{1}^{2}\right)\) (C) \(\frac{n}{2 n-1}\left(a_{1}^{2}-a_{2 n}^{2}\right)\) (D) None of these

Step-by-Step Solution

Answer

The answer is (D) None of these.

Step 1: Understand the Sequence

We have an arithmetic progression (AP) where each term follows the formula \( a_m = a_1 + (m-1)d \). Here, \( a_1 \) is the first term and \( d \) is the common difference.

Step 2: Identify the Series Pattern

The series given is of the form: \( a_1^2 - a_2^2 + a_3^2 - a_4^2 + \, \cdots \, + a_{2n-1}^2 - a_{2n}^2 \). Notice that each pair \( a_{2k-1}^2 - a_{2k}^2 \) can be written as a difference of squares.

Step 3: Apply Difference of Squares

The difference of squares formula is \( a^2 - b^2 = (a-b)(a+b) \). Applying this to each term, \( a_{2k-1}^2 - a_{2k}^2 = (a_{2k-1} - a_{2k})(a_{2k-1} + a_{2k}) \).

Step 4: Simplify the Pairwise Differences

For an AP, \( a_{2k-1} = a_1 + (2k-2)d \) and \( a_{2k} = a_1 + (2k-1)d \). So, \( a_{2k-1} - a_{2k} = -d \) and \( a_{2k-1} + a_{2k} = 2a_1 + (4k-3)d \).

Step 5: Evaluate the Series Sum

Thus, the entire series becomes: \[ \sum_{k=1}^{n} (a_{2k-1}^2 - a_{2k}^2) = \sum_{k=1}^{n} \, -d (2a_1 + (4k-3)d) \].

Step 6: Separate and Calculate Each Part

The sum is \(-d \sum_{k=1}^{n} (2a_1 + (4k-3)d)\). Expanding this: \[ (\text{Sum}_1) = -d \cdot 2a_1 \cdot n \] \[ (\text{Sum}_2) = -d \cdot d \sum_{k=1}^{n} (4k-3) \].

Step 7: Progress Through Arithmetic Series Formulas

The sum \( \sum_{k=1}^{n} 4k = 4 \sum_{k=1}^{n} k = 4 \cdot \frac{n(n+1)}{2} = 2n(n+1) \) and \( \sum_{k=1}^{n} 3 = 3n \) so \( \sum_{k=1}^{n} (4k-3) = 2n^2 + n \).

Step 8: Substitute Back and Simplify

Substitute back into the expression: \[ (\text{Sum}_2) = -d^2 (2n^2 + n) \]. Combine both parts: \[ -2a_1dn - d^2(2n^2 + n) \].

Step 9: Relate to Possible Answers

Given the complex result relates to the provided options, you see if any are simplified versions or transformations of this result. Option (D) is closest since other options don't reflect a clear simplified expression of the derived formula.

Key Concepts

Difference of Squares

The difference of squares is a mathematical concept that simplifies expressions where two squares are subtracted from each other. The identity is formulated as \( a^2 - b^2 = (a-b)(a+b) \). This allows us to factor the expression into a product of two simpler terms.

• The first term \((a-b)\) represents the difference between the two original terms \(a\) and \(b\).
• The second term \((a+b)\) is their sum.

This identity is extremely useful in simplifying complex algebraic expressions, especially when dealing with sequences or series.

In our exercise, each term in the series like \(a_{1}^2 - a_{2}^2\) can be expressed as the difference of squares. By applying the formula \((a_{2k-1} - a_{2k})(a_{2k-1} + a_{2k})\), we simplify the pair and proceed to solve the series more effectively.
It is important to recognize where this rule can make calculations easier, especially in the context of solving larger expressions.

Series Summation

Series summation is the process of adding up terms in a series. For arithmetic series, where each term increases by a constant amount, this process is well defined. Whether handling just a few or numerous terms, series summation helps in reaching a concise result.

To find the summation of the

• form \(a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots\), you begin by considering how each term interacts.
• In our exercise, once each pair is simplified using the difference of squares, we sum the results over \(n\) terms.

The goal often involves creating a compact expression that covers all terms comprehensively.
In this specific context, this involved each pair contributing a negative component—highlighted as \(-d (2a_1 + (4k-3)d)\). Summing over all \(n\) pairs gives a complete result of the series, which is vital for comparison with given answer options.

Common Difference

Within any Arithmetic Progression (AP), the common difference refers to the fixed number added to each term to arrive at the next term. Denoted often by \(d\), it's a core feature characterizing the progression.

In the given exercise, the common difference influences the pairwise terms \(a_{2k-1}\) and \(a_{2k}\).

• Each term is calculated as \( a_m = a_1 + (m-1)d \).
• The expression \(a_{2k-1} - a_{2k} = -d\) reflects the negative impact of the common difference between consecutive terms.

Recognizing the role of the common difference is key to understanding how terms are related and how the resulting series behaves.
It’s critical when simplifying expressions in arithmetic progressions, allowing for the calculation of partial sums and fitting series results to expected outcomes or models.