Finding Equal Terms In Arithmetic Progressions A Step-by-Step Guide

Jul 12, 2025 by ADMIN 68 views

Finding the Equal Terms in Two Arithmetic Progressions

In the realm of mathematics, arithmetic progressions (APs) hold a fundamental position. An arithmetic progression, simply put, is a sequence of numbers where the difference between any two consecutive terms remains constant. This constant difference is known as the common difference. Exploring the properties and relationships within and between different arithmetic progressions can lead to fascinating mathematical insights. In this article, we delve into a specific problem involving two arithmetic progressions and aim to find the value of n for which their nth terms are equal. We will meticulously walk through the steps, providing explanations and insights along the way, so you will deeply understand the process of solving this type of problem.

Understanding Arithmetic Progressions

Before we tackle the problem directly, let's solidify our understanding of arithmetic progressions. An arithmetic progression is defined by its first term (a) and its common difference (d). The general form of an AP is:

a, a + d, a + 2d, a + 3d, ...

Where:

a is the first term.
d is the common difference.

The nth term of an arithmetic progression, denoted as a_n, can be calculated using the formula:

a_n = a + (n - 1)d

This formula is crucial for solving problems related to arithmetic progressions, as it allows us to find any term in the sequence without having to list out all the preceding terms. Understanding this formula is key to mastering arithmetic progressions and solving related problems effectively. It's the bedrock upon which we will build our solution to the main problem at hand. Let's remember that the beauty of mathematics often lies in its ability to provide clear and concise formulas that help us make complex calculations easier and more intuitive.

Problem Statement

Now, let's formally state the problem we aim to solve. We are given two arithmetic progressions:

63, 65, 67, ...
3, 10, 17, ...

The question asks: For what value of n are the nth terms of these two arithmetic progressions equal? In other words, we need to find the position (n) in the sequences where the terms have the same value. This involves determining the general form of each AP, setting their nth term expressions equal to each other, and then solving for n. It's a classic problem that combines the principles of arithmetic progressions with basic algebraic techniques. Solving this problem will not only give us a numerical answer but also a deeper insight into the behavior and properties of arithmetic sequences. This problem underscores the power of mathematical tools in uncovering relationships between seemingly different sequences.

Identifying the Parameters of the Arithmetic Progressions

To find the nth terms of the given arithmetic progressions, we first need to identify their respective first terms (a) and common differences (d). Let's analyze each AP separately.

Arithmetic Progression 1: 63, 65, 67, ...

The first term, a₁, is 63.
The common difference, d₁, is the difference between any two consecutive terms. For example, 65 - 63 = 2. So, d₁ = 2.

Arithmetic Progression 2: 3, 10, 17, ...

The first term, a₂, is 3.
The common difference, d₂, is the difference between any two consecutive terms. For example, 10 - 3 = 7. So, d₂ = 7.

Identifying these parameters is a critical first step. We now have the necessary information to express the nth terms of both APs in a general form. This is like setting the stage for the core of our solution, where we equate the two expressions and solve for the unknown variable, n. The accurate determination of these parameters ensures that our subsequent calculations are grounded in the correct values, leading us to a precise and reliable answer. It also highlights the systematic approach in solving mathematical problems – break them down into smaller, manageable parts, and then address each part methodically.

Expressing the nth Terms

Now that we have identified the first terms and common differences, we can express the nth term for each arithmetic progression using the formula: a_n = a + (n - 1)d.

nth Term of Arithmetic Progression 1

Using the values a₁ = 63 and d₁ = 2, the nth term, denoted as a_n1, is:

a_n1 = 63 + (n - 1) * 2
a_n1 = 63 + 2n - 2
a_n1 = 2n + 61

nth Term of Arithmetic Progression 2

Using the values a₂ = 3 and d₂ = 7, the nth term, denoted as a_n2, is:

a_n2 = 3 + (n - 1) * 7
a_n2 = 3 + 7n - 7
a_n2 = 7n - 4

We now have expressions for the nth terms of both arithmetic progressions. These expressions are the key to finding the value of n for which the terms are equal. It's like having two different routes that lead to the same destination, and our goal is to find where those routes intersect. The process of expressing the nth terms in this algebraic form allows us to directly compare and solve for the unknown, rather than manually listing out terms until we find a match, which would be an impractical approach for larger sequences.

Equating the nth Terms and Solving for n

To find the value of n for which the nth terms of the two arithmetic progressions are equal, we set the expressions for a_n1 and a_n2 equal to each other:

2n + 61 = 7n - 4

Now, we solve this equation for n. This involves rearranging the terms to isolate n on one side of the equation.

Subtract 2n from both sides:
- 61 = 5n - 4
Add 4 to both sides:
- 65 = 5n
Divide both sides by 5:
- n = 13

Therefore, the nth terms of the two arithmetic progressions are equal when n = 13. This is our solution. We have successfully found the position in the sequences where the terms have the same value. The algebraic manipulation we've done here is a classic example of how equations can be used to solve real-world problems or abstract mathematical inquiries. The beauty of this approach is that it's precise and efficient, providing us with the answer without the need for trial and error.

Verification

To ensure our solution is correct, we can substitute n = 13 into the expressions for a_n1 and a_n2 and verify that they yield the same value.

For Arithmetic Progression 1:

a_13,1 = 2 * 13 + 61
a_13,1 = 26 + 61
a_13,1 = 87

For Arithmetic Progression 2:

a_13,2 = 7 * 13 - 4
a_13,2 = 91 - 4
a_13,2 = 87

Since a_13,1 = a_13,2 = 87, our solution is verified. This verification step is a crucial part of the problem-solving process. It's like double-checking your work in any important task, ensuring that you haven't made any errors along the way. It gives us confidence in our answer and demonstrates the consistency and reliability of the mathematical methods we've used. In this case, the fact that both expressions yield the same value when n = 13 confirms that our algebraic manipulations and the final result are accurate.

Conclusion

In conclusion, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... are equal when n = 13. We arrived at this solution by first understanding the concept of arithmetic progressions, then identifying the first terms and common differences of the given sequences. We then expressed the nth terms of both APs using the general formula a_n = a + (n - 1)d. By equating these expressions and solving for n, we found the desired value. Finally, we verified our solution by substituting n = 13 back into the expressions and confirming that they yielded the same term. This problem demonstrates the power of algebraic techniques in solving problems related to arithmetic progressions and highlights the importance of a systematic approach to problem-solving in mathematics. The process of solving this problem is a microcosm of the larger mathematical endeavor – a combination of understanding fundamental concepts, applying appropriate formulas, performing accurate calculations, and verifying the results.