Problem 25 - Beginning and Intermediate Algebra

Question

For each arithmetic sequence, find \(a_{n}\) and then use \(a_{n}\) to find the indicated term. $$-7,-5,-3,-1,1, \ldots ; a_{25}$$

Step-by-Step Solution

Answer

The general term of the arithmetic sequence is \(a_{n} = -7 + (n - 1)2\). Using this formula, we can find the 25th term, \(a_{25} = -7 + (25 - 1)2 = -7 + 48 = 41\). So, the 25th term of the sequence is 41.

Step 1: Determine the Common Difference

To find the common difference, subtract the current term from the next term in the sequence. In our case, the first few terms are \(-7, -5, -3, -1\), so the common difference is: $$-5 - (-7) = -5 + 7 = 2$$ So, the common difference between the terms of the sequence is 2.

Step 2: Find the Formula for the nth Term \(a_{n}\)

An arithmetic sequence can be written using the general formula: $$a_{n} = a_{1} + (n-1)d$$ where \(a_{n}\) is the nth term, \(a_{1}\) is the first term, \(n\) is the position of the term, and \(d\) is the common difference. In our case, we have the first term \(a_{1}=-7\) and the common difference \(d=2\). So, the formula for the nth term is: $$a_{n} = -7 + (n-1)2$$

Step 3: Find the 25th Term of the Sequence \(a_{25}\)

Now, we want to find the 25th term of the sequence, so we plug in \(n=25\) into the formula \(a_{n} = -7 + (n - 1)2\): $$a_{25} = -7 + (25 - 1)2$$ Calculating the expression, we get: $$a_{25} = -7 + (24)(2) = -7 + 48 = 41$$ The 25th term of the arithmetic sequence is 41.

Key Concepts

Common Difference

In an arithmetic sequence, the common difference is a fundamental concept that helps define the pattern of the sequence. It is the constant amount that each term increases or decreases by as you move from one term to the next. To find the common difference, you subtract any term in the sequence from the term that follows it.
For example, in the sequence \(-7, -5, -3, -1, 1, \ldots\), observe the difference between the first two terms:

• \(-5 - (-7) = -5 + 7 = 2\)

This calculation shows that the common difference \(d\) is \(2\). It tells us that every term is \(2\) more than the previous one.
Understanding the common difference is crucial as it is used in formulas to determine other terms within the sequence.

nth Term Formula

The nth term formula of an arithmetic sequence is invaluable for finding any specific term without listing all previous terms. The general formula is:

• \(a_n = a_1 + (n-1)d\)

where:

• \(a_n\) is the n'th term,
• \(a_1\) is the first term,
• \(n\) is the term's position in the sequence, and
• \(d\) is the common difference.

Using the sequence from our example \(-7, -5, -3, -1, 1, \ldots\), the first term \(a_1\) is \(-7\) and the common difference \(d\) is \(2\). Applying these values to the nth term formula, we get:

• \(a_n = -7 + (n-1)2\)

This formula allows you to find any term in the sequence directly.

25th Term

Finding the 25th term in an arithmetic sequence involves substituting \(n = 25\) into the nth term formula. With our sequence formula \(a_n = -7 + (n-1)2\), the task becomes straightforward:

• \(a_{25} = -7 + (25-1)2\)

Break it down into calculations:

• First, \(25 - 1 = 24\)
• Next, multiply by the common difference: \(24 \times 2 = 48\)
• Add this to the first term: \(-7 + 48 = 41\)

So, the 25th term \(a_{25}\) is \(41\). This process can be used to find any term in the sequence effortlessly.

Sequence Formula

The sequence formula is a critical tool for understanding and describing patterns in arithmetic sequences. It provides a way to calculate any term within the sequence by using a structured formula:
The sequence formula for an arithmetic sequence is:

• \(a_n = a_1 + (n-1)d\)

This formula incorporates:

• The first term \(a_1\), which is the starting point of the sequence,
• The common difference \(d\), determining how much we add to each term to get the next one,
• The position \(n\), indicating which term's value we are calculating.

Understanding this formula is vital because it allows for quick calculations of any term's value, rather than sequentially computing each preceding term. This efficiency makes arithmetic sequences manageable and predictable.