Problem 53 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The number of terms of an A.P. is even; the sum of the odd terms is 24 , of the even terms 30 , and the last term exceeds the first by \(10 \frac{1}{2}\), find the number of terms and the series.

Step-by-Step Solution

Answer

The number of terms is 8, and the series is 6.5, 9.5, 12.5, 15.5, 18.5, 21.5, 24.5, 27.5.

Step 1: Determine basic formula

The formula for the sum of an arithmetic series is \[ S = \frac{n}{2} (a + l) \], where \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term of the arithmetic series.

Step 2: Determine first two terms

The sum of the odd terms and even terms, \( S_{odd} = 24 \) and \( S_{even} = 30 \) provide us two equations. As \(n\) is even, the number of odd terms equals the number of even terms (\(n/2\)). We can represent the odd terms as \(a, a+2d, a+4d, ...\) and even terms as \(a+d, a+3d, a+5d, ...\), where \(d\) is the common difference of arithmetic progression. Thus, we get the equations: \( \frac{n}{2}(2a + (n-1)d) = 24 \) and \( \frac{n}{2}(2a + 2d + (n-1)d) = 30 \). From these, we can find the two values of \(a + \frac{(n-1)d}{2}\), which are 12 and 15.

Step 3: Find common difference and first term

Subtract the first value from the second to get the common difference: \(d = 15 - 12 = 3\). Then we find the first term by using the given condition that the last term exceeds the first by \(10 \frac{1}{2}\). So using \(l = a + (n-1)d\), and substituting \(d\), \(a\) and \(l\), we get \(a = \frac{21 - n}{2}\). Substituting this in our equation from step 2, we get \(n=8\).

Step 4: Determine the series

The first term is \(a = \frac{21 - n}{2} = \frac{13}{2} = 6.5\). The whole series thus becomes \(6.5, 9.5, 12.5, 15.5, 18.5, 21.5, 24.5, 27.5\).

Key Concepts

Sum of arithmetic series

An arithmetic progression (A.P.) is a sequence of numbers where each term after the first is obtained by adding a fixed number, called the common difference, to the previous term. Understanding the sum of such a series is essential in solving many mathematical problems.

To find the sum of an arithmetic series, you can use the formula:

• \[ S = \frac{n}{2} (a + l) \]

where:

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

This formula helps to quickly calculate the total of all terms in an arithmetic series without adding each term individually.

In our specific problem, this formula was used to determine both the sum of odd and even terms separately. Such calculations often help in identifying key features of the sequence like the number of terms or the values of individual terms.

Common difference in AP

In an arithmetic progression, the common difference (\(d\)) is crucial as it defines the sequence pattern. It is the difference between any two successive terms, so you can calculate it by subtracting any term from the next term in the sequence.

For our problem, the common difference was found using the relationship:

• \(a + (n-1)d = l\),

where \(l\) is the last term in the series. Given that the last term exceeds the first by \(10\frac{1}{2}\), we were able to deduce that the common difference \(d\) is 3 by subtracting the sum components from the respective odd and even sum equations.

Finding the common difference allowed us to explore further and determine the values of other terms in the series, eventually leading to the discovery of the complete sequence.

Terms of an arithmetic progression

The terms of an arithmetic progression follow a defined pattern, given by the formula for any term (\(T_n\)):

• \(T_n = a + (n-1)d\)

This formula helps to find the value of each term based on its position \(n\), the first term \(a\), and the common difference \(d\).

In the given exercise, the focus was on understanding how both odd and even terms fit into the complete series. The terms were split into odd and even sequences based on their position in the sequence, using the first term and the common difference calculated earlier.

For example, the first few terms could be calculated as:

• Odd terms: \(a, a+2d, a+4d, ...\)
• Even terms: \(a+d, a+3d, a+5d, ...\)

This breakdown allows for finding any individual term needed in problem solving and confirms the structure of the series itself.