Problem 72 - College Algebra with Corequisite Support

Question

For which terms does the finite arithmetic sequence \(\left\\{\frac{5}{2}, \frac{19}{8}, \frac{9}{4}, \ldots, \frac{1}{8}\right\\}\) have integer values?

Step-by-Step Solution

Answer

The sequence has integer terms for \( n = 21, 13, \) and \( 5 \).

Step 1: Understand the sequence pattern

The given sequence is arithmetic, which means there is a constant difference between consecutive terms. The sequence starts with \( \frac{5}{2} \) and the next term is \( \frac{19}{8} \). We need to find the common difference \( d \).

Step 2: Calculate the common difference

The common difference \( d \) can be calculated by subtracting the first term from the second term: \( d = \frac{19}{8} - \frac{5}{2} \). Convert \( \frac{5}{2} \) to \( \frac{20}{8} \) to have a common denominator. Thus, \( d = \frac{19}{8} - \frac{20}{8} = -\frac{1}{8} \).

Step 3: Write the formula for the nth term

In an arithmetic sequence, the \( n \)-th term \( a_n \) is given by \( a_n = a_1 + (n-1)d \). Here, \( a_1 = \frac{5}{2} \) and \( d = -\frac{1}{8} \). So, the \( n \)-th term is \( a_n = \frac{5}{2} + (n-1) \left(-\frac{1}{8}\right) \).

Step 4: Simplify the formula for the nth term

Simplifying, \( a_n = \frac{5}{2} - \frac{n-1}{8} \). Convert \( \frac{5}{2} \) to \( \frac{20}{8} \) for common denominators: \( a_n = \frac{20}{8} - \frac{n-1}{8} = \frac{20 - (n-1)}{8} = \frac{21-n}{8} \).

Step 5: Determine when the nth term is an integer

The term \( \frac{21-n}{8} \) is an integer when \( 21-n \) is a multiple of 8. Let \( 21-n = 8k \) for some integer \( k \). This implies \( n = 21 - 8k \).

Step 6: Solve for possible integer terms

Substitute plausible integer values of \( k \) so that \( n \) remains positive. For integer \( n \), and considering the starting term \( \frac{5}{2} \), \( k = 0, 1, 2 \) gives \( n = 21, 13, 5 \) respectively. These are the terms where \( \frac{21-n}{8} \) is an integer.

Key Concepts

Common Difference

In an arithmetic sequence, the concept of a common difference is fundamental. It is the constant value you add or subtract to move from one term to the next. This difference is crucial because it defines the pattern of the sequence. To find the common difference, you simply subtract any term from the term that follows it. For example, in our sequence of \( \left\{\frac{5}{2}, \frac{19}{8}, \frac{9}{4}, \ldots, \frac{1}{8}\right\} \), the common difference \( d \) is found by subtracting the first term from the second. We perform the calculation: - Convert \( \frac{5}{2} \) to \( \frac{20}{8} \), so both fractions have the same denominator.- Then, subtract \( \frac{19}{8} - \frac{20}{8} = -\frac{1}{8} \). Now, we know the common difference is \(-\frac{1}{8}\), meaning each subsequent term is \(\frac{1}{8}\) less than the previous term. This consistent pattern makes it easier to predict other terms in the sequence.

nth Term Formula

The nth term formula is a way to directly find any term in the sequence without listing all terms prior to it. It's a critical tool in working through arithmetic sequences because it allows us to calculate any term using its position in the sequence. The formula for the nth term \( a_n \) in an arithmetic sequence is given by:\[ a_n = a_1 + (n-1)d \]where:

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

Given our sequence, we have \( a_1 = \frac{5}{2} \) and \( d = -\frac{1}{8} \). Plug these values into the formula:- \( a_n = \frac{5}{2} + (n-1)(-\frac{1}{8}) \)- Simplifying: \( a_n = \frac{5}{2} - \frac{n-1}{8} \).- Convert to a common denominator: \( \frac{5}{2} = \frac{20}{8} \), resulting in \( a_n = \frac{20 - (n-1)}{8} \)- Final form: \( a_n = \frac{21-n}{8} \).This simplified formula allows for easy calculation of any term \( a_n \) when \( n \) is specified.

Integer Values

To find when the terms of an arithmetic sequence are integers, you need the sequence expression to yield whole numbers. For our sequence, expressed by the formula \( \frac{21-n}{8} \), the condition for the term to be an integer is that the numerator \( 21-n \) should be divisible by 8.We express this requirement mathematically:- Let \( 21-n = 8k \), where \( k \) is any integer, which results in \( n = 21 - 8k \).Next, we determine integer values of \( n \) by substituting possible integer values for \( k \) such that \( n \) remains a positive integer:

• For \( k = 0 \), \( n = 21 \)
• For \( k = 1 \), \( n = 13 \)
• For \( k = 2 \), \( n = 5 \)

These values of \( n \) indicate the positions in the sequence where the terms will be integers. These calculations help ensure we understand exactly where integer terms appear within our sequence. Knowing how to discern this helps tremendously, especially when dealing with sequences comprising fractions.