Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The first and last term of an A.P. are \(a\) and \(l\) respectively. If \(S\) is the sum of all the terms of the A.P. and the common difference is \(\frac{l^{2}-a^{2}}{k-(l+a)}\), then \(k\) is equal to (A) \(S\) (B) \(2 S\) (C) \(3 S\) (D) None of these

Step-by-Step Solution

Answer

(B) 2S

Step 1: Understand the given terms

We have an arithmetic progression (A.P.) with first term \(a\) and last term \(l\). The sum of all terms is given as \(S\), and the common difference is \( \frac{l^2 - a^2}{k - (l+a)} \). The task is to find the relationship of \(k\) with \(S\).

Step 2: Recall A.P. sum formula

The sum \(S\) of \(n\) terms in an A.P. with first term \(a\), last term \(l\) is given by:\[ S = \frac{n}{2} (a + l) \]

Step 3: Apply common difference formula

The common difference \(d\) of the A.P. can also be expressed using the terms:\[ d = \frac{l - a}{n - 1} \]Given, another expression for \(d\) is \( \frac{l^2 - a^2}{k - (l+a)} \).

Step 4: Equate the expressions for common difference

Set the two expressions for \(d\) equal to each other:\[ \frac{l-a}{n-1} = \frac{l^2 - a^2}{k - (l+a)} \]

Step 5: Simplify the equation

Cross-multiplying gives:\[ (l-a)(k-(l+a)) = (n-1)(l^2 - a^2) \]

Step 6: Expand and simplify

Expand and simplify the left side:\[ lk - l^2 - la + ak + a^2 + al = (n-1)(l+a)(l-a) \]Which simplifies to\[ (l-a)k = (n-1)(l^2 - a^2) \] after cancelling terms.

Step 7: Use the identity to solve for \(k\)

Using the identity \(l^2 - a^2 = (l-a)(l+a)\):\[ (l-a)k = (n-1)(l-a)(l+a) \] Solving for \(k\) gives:\[ k = (n-1)(l+a) \]

Step 8: Substitute \((n-1)(l+a)\) in terms of \(S\)

Recall from Step 2, \(n = \frac{2S}{a+l}\), thus:\[ k = \frac{2S}{a+l} \cdot (l+a) = 2S \]

Step 9: Conclude the solution

Therefore, \(k\) is equivalent to \(2S\).

Key Concepts

Sum of Arithmetic Series

The sum of an arithmetic series is a key concept in understanding arithmetic progressions. An arithmetic progression, or A.P., is a sequence of numbers in which each term after the first is the sum of the previous term and a constant, called the common difference. When calculating the sum of the first "n" terms of an A.P., you can use the formula:

• \( S = \frac{n}{2} (a + l) \)

Here, \(S\) is the sum of the series, \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term of the series.

The idea is straightforward: you multiply the average of the first and the last term by the number of terms. This formula sums up all terms between the first and the last in the sequence, making it efficient for calculating series where terms follow this specific repetitive order. The simplicity of this formula allows for quick calculations in various arithmetic series scenarios.

Common Difference in A.P.

The common difference in an arithmetic progression is the value added to each term to get the following term. Understanding this concept is essential as it helps to define the entire series. You calculate this by subtracting the first term from the second term or using the formula:

• \( d = \frac{l - a}{n - 1} \)

where \(d\) represents the common difference, \(l\) is the last term, \(a\) is the first term, and \(n\) is the number of terms.

Different formulas for the common difference can be derived using algebraic manipulation. For instance, in this exercise, another expression for the common difference was \(\frac{l^2 - a^2}{k - (l+a)}\). By setting these formulas equal, you can discover additional relationships between variables in arithmetic progression problems. Recognizing these relationships can help solve complex algebraic puzzles and is particularly useful in exercises involving finding unknowns.

Algebraic Manipulation

Algebraic manipulation is a method of transforming equations to reveal needed variables or relationships. This technique involves rearranging and simplifying equations using basic algebraic skills.

In the provided exercise, one key step involved equating different expressions for the common difference, showing:

• \( \frac{l-a}{n-1} = \frac{l^2 - a^2}{k - (l+a)} \)

From there, cross-multiplying and simplifying the equation allowed for further exploration of the relationship between \(k\) and \(S\). The ability to easily manipulate these equations is crucial as it helps unravel the mystery behind unknowns like \(k\).

Through algebraic expansion, multiplication, and simplification, complex arithmetic problems become more manageable. Practicing these techniques allows students to improve their problem-solving skills and adapt to various mathematical challenges.