Problem 14 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(a_{1}, a_{2}, a_{3}, \ldots . . a_{n}\) are in A.P. and \(a_{1}+a_{4}+a_{7}+\ldots+a_{16}=114\) then \(a_{1}+a_{6}+a_{11}+a_{16}\) is equal to : \(\quad\) April 10,2019 (I)] (a) 98 (b) 76 (c) 38 (d) 64

Step-by-Step Solution

Answer

76

Step 1: Understand the Problem

We are given an arithmetic progression (A.P.) with terms \(a_1, a_2, a_3, \ldots, a_n\). The sum of certain terms, specifically \(a_1, a_4, a_7, \ldots, a_{16}\), is 114. We need to find the sum of the terms \(a_1, a_6, a_{11}, a_{16}\).

Step 2: Identify the Pattern of Terms

The sequence of terms \(a_1, a_4, a_7, \ldots, a_{16}\) suggests that each term can be expressed as \(a_{1 + 3(k-1)}\) for integer \(k\). The terms \(a_1, a_6, a_{11}, a_{16}\) can be expressed as \(a_{1 + 5(m-1)}\) for integer \(m\).

Step 3: Use the General Term of an A.P.

In an arithmetic progression, the general term can be written as \(a_n = a_1 + (n-1)d\), where \(d\) is the common difference. For instance, \(a_4 = a_1 + 3d\) for the sequence starting from \(a_1\), with steps of 3 in-between terms.

Step 4: Calculate the First Sum

The terms summed are \(a_1 + (a_1 + 3d) + (a_1 + 6d) + \ldots + (a_1 + 15d)\). This is an A.P. with first term \(a_1\) and common difference \(3d\). When rearranging this, the total number of terms is 6, as the sequence is \(a_1, a_4, a_7, a_{10}, a_{13}, a_{16}\). The sum can be calculated using: \[ S = \frac{n}{2} [2a_1 + (n-1)3d] \] where \(n = 6\). This sum is given as 114.

Step 5: Solve for Unknowns

Now solve for \(a_1\) and \(d\): \[ 114 = \frac{6}{2}[2a_1 + 15d] = 3[2a_1 + 15d] \] Simplifying, we get: \[ 2a_1 + 15d = 38 \]

Step 6: Calculate the Second Sum

Using the same approach as before, we calculate the sum of \(a_1, a_6, a_{11}, a_{16}\), which are characterized by terms \(a_1, a_1 + 5d, a_1 + 10d, a_1 + 15d\). This forms another A.P. with first term \(a_1\) and common difference \(5d\), and the number of terms is 4. The sum is: \[ S_2 = \frac{4}{2} [2a_1 + 15d] = 2[2a_1 + 15d] = 76 \]

Step 7: Conclusion

The problem asks for \(a_1 + a_6 + a_{11} + a_{16}\), and from the calculations, we find this sum to be equal to 76.

Key Concepts

Sum of Arithmetic Series

An arithmetic series is a sum of the terms in an arithmetic progression (A.P.). In an A.P., every term after the first is formed by adding a constant, called the common difference, to the previous term. For an arithmetic series, the sum of the terms can be calculated using a specific formula. This formula is handy because it allows us to quickly find out how much the terms add up to without manually summing each of them.

For a series with the first term as \(a_1\), the number of terms \(n\), and the common difference \(d\), the formula is given by:

• \[ S = \frac{n}{2} \left( 2a_1 + (n-1)d \right) \]

In this formula:

• \(n\) represents the number of terms in the series.
• \(a_1\) is the first term of the series.
• \(d\) is the common difference between consecutive terms.

This formula efficiently gives us the total of the series by leveraging the pattern in which terms change. It's important to remember that this sum can only be exact when the sequence follows the A.P. pattern.

Common Difference

The common difference in an arithmetic progression (A.P.) is the interval between each term. It remains the same throughout the entire series, which is what makes an arithmetic sequence predictable.

To find the common difference \(d\), you simply subtract any term from the next term in the sequence. For example, in the series \(a_1, a_2, a_3,\ldots\), the common difference \(d\) is \(a_2 - a_1\).

Why is the common difference significant?

• It is used to derive the general term of the sequence.
• It helps determine the rate at which the sequence outcomes progress.

By maintaining a fixed difference between terms, arithmetic progressions provide a reliable way to calculate specific terms and sums within the sequence.

General Term of Sequence

In an arithmetic sequence, the general term formula is crucial for finding any specific term. This formula allows you to express the \(n\)th term, \(a_n\), of an arithmetic sequence using the first term and the common difference.

The general term \(a_n\) can be written as:

• \[ a_n = a_1 + (n-1)d \]

Here:

• \(a_1\) is the first term of the sequence.
• \(n\) is the position of the term in the sequence.
• \(d\) is the common difference.

This formula helps quickly determine any term within the arithmetic progression without having to list all the terms leading up to it. By inserting the values for \(a_1\), \(d\), and \(n\), we can easily compute the desired term's value, streamlining arithmetic progression calculations.