Problem 32 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots .\) are in A.P. such that \(a_{4}-a_{7}+a_{10}=m\), then the sum of first 13 terms of this A.P., is: [Online April 23, 2013] (a) \(10 \mathrm{~m}\) (b) \(12 \mathrm{~m}\) (c) \(13 \mathrm{~m}\) (d) \(15 \mathrm{~m}\)

Step-by-Step Solution

Answer

The sum of the first 13 terms is \(13m\).

Step 1: Understand the Problem

We have an arithmetic progression (A.P.) given by terms \(a_1, a_2, a_3, \ldots\). The formula for the general term of an A.P. is \(a_n = a_1 + (n-1)d\), where \(d\) is the common difference. We are given that \(a_4 - a_7 + a_{10} = m\). We need to find the sum of the first 13 terms of this A.P., denoted as \(S_{13}\).

Step 2: Write Specific Terms in Terms of \(a_1\) and \(d\)

Using the formula for the general term of an A.P., express the specific terms: - \(a_4 = a_1 + 3d\)- \(a_7 = a_1 + 6d\)- \(a_{10} = a_1 + 9d\)

Step 3: Set up the Given Equation

Substitute the expressions for \(a_4, a_7, \) and \(a_{10}\) into the condition given in the problem:\((a_1 + 3d) - (a_1 + 6d) + (a_1 + 9d) = m\)

Step 4: Simplify the Given Equation

Simplify the equation:\[(a_1 + 3d) - (a_1 + 6d) + (a_1 + 9d) = a_1 + 3d - a_1 - 6d + a_1 + 9d\]\[= a_1 + 6d = m\]

Step 5: Find the Sum of the First 13 Terms

The formula for the sum of the first \(n\) terms of an A.P. is given by:\[S_n = \frac{n}{2}(2a_1 + (n-1)d)\]For 13 terms, substitute \(n = 13\) and simplify using \(2a_1 + 12d\):\[S_{13} = \frac{13}{2}(2a_1 + 12d)\] Since \(a_1 + 6d = m\), we have \(2a_1 + 12d = 2m\).Thus, substitute \(2m\) in place of \(2a_1 + 12d\):\[S_{13} = \frac{13}{2} \times 2m = 13m\]

Step 6: Select the Correct Answer

The result shows that the sum of the first 13 terms of the A.P. is \(13m\). Referring to the options, (c) \(13m\) is the correct answer.

Key Concepts

Sum of Terms

In an arithmetic progression (A.P.), the sum of the terms up to a certain point is a common requirement you will encounter. The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \]
Here, \( S_n \) represents the sum of the first \( n \) terms, \( a_1 \) is the first term, \( n \) is the number of terms, and \( d \) is the common difference between successive terms.

• The formula essentially calculates the sum by averaging the first and last terms and then multiplying by the number of terms.
• It efficiently sums the series without needing to add each term individually.

Using this formula is particularly useful because you can quickly adjust it to find the sum of any number of sequential terms in a progression.

Common Difference

In an arithmetic progression, the 'common difference' \( d \) is what separates each term from the next. It is the uniform amount you add (or subtract if negative) to get from one term to the next. Determining this difference is crucial because it informs how the terms in the sequence relate to one another.
To find the common difference in a sequence:

• Subtract the first term from the second term, \( d = a_2 - a_1 \).
• This subtraction can be applied to any two successive terms: \( d = a_{n+1} - a_n \).

Understanding the common difference allows you to predict future terms and understand the general structure of the sequence. The constant nature of \( d \) is what makes the sequence truly arithmetic.

General Term Formula

The general term formula of an arithmetic progression helps determine any specific term in the sequence. This is vital for problems where you might need to isolate or work with a particular term in progression.
The formula is:
\[ a_n = a_1 + (n-1)d \]
Here, \( a_n \) is the term you want to find, \( a_1 \) is the first term of the sequence, \( n \) is the position of the term in the sequence, and \( d \) is the common difference.

• This formula shows that every term is constructed by starting from the first term and adding the common difference \( (n-1) \) times.
• It's a crucial tool for means testing or verifying specific terms, especially in proofs or larger calculations.

Series and Sequences

A sequence is an ordered list of numbers, each of which is called a term. In the context of arithmetic progressions, each term is derived by adding a common difference to the previous term. When we talk about a 'series,' we are considering the sum of several terms of a sequence.

• An arithmetic sequence progresses by a consistent amount called the common difference, forming a linear pattern.
• When the terms of a sequence are summed, it becomes a series.

Understanding sequences and series is central to solving problems in arithmetic progressions because each element and its sum relate back to the concepts of order, repetition, and addition.
These concepts can be extended beyond arithmetic to geometric sequences and series, where terms are multiplied by a fixed number, or to other mathematical contexts as well.