Finding Terms In Arithmetic Sequences Determining The 25th And 17th Term
In the realm of mathematics, sequences play a fundamental role, providing a structured way to represent ordered lists of numbers. Among these sequences, arithmetic sequences hold a special place due to their consistent and predictable nature. In this comprehensive exploration, we will delve into the fascinating world of arithmetic sequences, unraveling their underlying principles and applying them to solve specific problems. We will tackle the challenge of finding the 25th term of the sequence 7, 13, 19, ..., determine the 17th term of the sequence 15, 23, 31, ..., and investigate whether the number 239 is a member of this particular sequence. By the end of this journey, you will possess a solid understanding of arithmetic sequences and the tools to confidently navigate their intricacies.
Arithmetic sequences are characterized by a constant difference between consecutive terms. This constant difference, aptly named the common difference, forms the backbone of the sequence, dictating its progression. To embark on our exploration, let's first define the general form of an arithmetic sequence. An arithmetic sequence can be represented as follows:
a, a + d, a + 2d, a + 3d, ...
where:
- a represents the first term of the sequence.
- d denotes the common difference between consecutive terms.
The nth term of an arithmetic sequence, denoted as an, can be calculated using the following formula:
an = a + (n - 1)d
This formula is the cornerstone of our exploration, providing a direct link between the term number (n), the first term (a), the common difference (d), and the value of the nth term (an). With this formula in hand, we are well-equipped to tackle the challenges that lie ahead.
Finding the 25th Term of the Sequence 7, 13, 19, ...
Our first challenge involves determining the 25th term of the sequence 7, 13, 19, .... To conquer this challenge, we will employ the formula for the nth term of an arithmetic sequence. First, we need to identify the first term (a) and the common difference (d) of the sequence. By inspection, we can readily identify the first term as a = 7. To determine the common difference, we can subtract any term from its immediate successor. For instance, subtracting the first term (7) from the second term (13) yields a common difference of d = 13 - 7 = 6. Now that we have identified the first term (a = 7) and the common difference (d = 6), we can confidently apply the formula for the nth term to find the 25th term (a25).
a25 = a + (25 - 1)d
Substituting the values of a and d, we get:
a25 = 7 + (25 - 1)6
Simplifying the expression, we arrive at:
a25 = 7 + 24 * 6
a25 = 7 + 144
a25 = 151
Therefore, the 25th term of the sequence 7, 13, 19, ... is 151. This result showcases the power of the formula for the nth term, allowing us to efficiently calculate any term in the sequence without having to manually list out all the preceding terms.
Determining the 17th Term of the Sequence 15, 23, 31, ...
Our next task is to find the 17th term of the sequence 15, 23, 31, .... We will once again leverage the formula for the nth term of an arithmetic sequence. As before, we begin by identifying the first term (a) and the common difference (d). The first term is readily apparent as a = 15. To find the common difference, we subtract any term from its successor. Subtracting the first term (15) from the second term (23) gives us a common difference of d = 23 - 15 = 8. Now that we have the first term (a = 15) and the common difference (d = 8), we can apply the formula for the nth term to calculate the 17th term (a17).
a17 = a + (17 - 1)d
Substituting the values of a and d, we get:
a17 = 15 + (17 - 1)8
Simplifying the expression, we obtain:
a17 = 15 + 16 * 8
a17 = 15 + 128
a17 = 143
Therefore, the 17th term of the sequence 15, 23, 31, ... is 143. This result further demonstrates the utility of the formula for the nth term in efficiently determining specific terms within an arithmetic sequence.
Investigating the Membership of 239 in the Sequence 15, 23, 31, ...
Our final challenge involves investigating whether the number 239 is a term of the sequence 15, 23, 31, .... To address this question, we will employ a slightly different approach. We will assume that 239 is indeed a term of the sequence and then attempt to find the term number (n) that corresponds to this value. If we can find a positive integer value for n, then 239 is a member of the sequence. Otherwise, it is not.
Let's assume that 239 is the nth term of the sequence. Then, we can write:
239 = a + (n - 1)d
We already know that the first term (a) is 15 and the common difference (d) is 8. Substituting these values, we get:
239 = 15 + (n - 1)8
Now, we solve for n:
239 - 15 = (n - 1)8
224 = (n - 1)8
224 / 8 = n - 1
28 = n - 1
n = 28 + 1
n = 29
Since we have found a positive integer value for n (n = 29), we can conclude that 239 is indeed a term of the sequence 15, 23, 31, .... Specifically, it is the 29th term of the sequence. This result illustrates how we can use the formula for the nth term in reverse to determine whether a given number is a member of an arithmetic sequence.
In conclusion, we have successfully navigated the world of arithmetic sequences, employing the formula for the nth term to solve a variety of problems. We determined the 25th term of the sequence 7, 13, 19, ..., found the 17th term of the sequence 15, 23, 31, ..., and investigated whether 239 is a term of this latter sequence. These examples showcase the power and versatility of arithmetic sequences and their associated formulas. By mastering these concepts, you are well-equipped to tackle a wide range of mathematical challenges involving sequences and patterns. Remember, the key to success lies in understanding the underlying principles and applying the appropriate tools to solve the problem at hand. With practice and perseverance, you can unlock the secrets of arithmetic sequences and confidently navigate the world of mathematics.
