Problem 55 - College Algebra

Question

For the following exercises, find the number of terms in the given finite arithmetic sequence. $$ a=\left\\{\frac{1}{2}, 2, \frac{7}{2}, \ldots, 8\right\\} $$

Step-by-Step Solution

Answer

The sequence has 6 terms.

Step 1: Identify the First Term and Common Difference

The first term of the arithmetic sequence is \(a_1 = \frac{1}{2}\). To find the common difference \(d\), subtract the first term from the second term: \(d = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}\). So, the common difference is \(\frac{3}{2}\).

Step 2: Use the Formula for the nth Term

The formula for the nth term of an arithmetic sequence is \(a_n = a_1 + (n-1) \cdot d\). We know \(a_n = 8\), \(a_1 = \frac{1}{2}\), and \(d = \frac{3}{2}\). Substitute these values into the formula: \[8 = \frac{1}{2} + (n-1) \cdot \frac{3}{2}\]

Step 3: Solve for n

First, rewrite the equation to isolate \((n-1)\):\[8 - \frac{1}{2} = (n-1) \cdot \frac{3}{2}\]This simplifies to:\[\frac{15}{2} = (n-1) \cdot \frac{3}{2}\]Multiply both sides by \(\frac{2}{3}\):\[\frac{15}{2} \times \frac{2}{3} = n-1\]This results in:\[5 = n-1\] Finally, add 1 to both sides to solve for \(n\):\[n = 6\]

Step 4: Conclude the Number of Terms

The finite arithmetic sequence has \(6\) terms.

Key Concepts

Common Difference

In arithmetic sequences, the common difference is the consistent interval that separates sequential terms. It's the defining feature of this type of sequence. Imagine you're walking up a gentle hill, where each step represents the transition from one term to the next. The distance between each step is your common difference. To find it, simply subtract the first term from the second term in the sequence. For example, if we're looking at the sequence \(\frac{1}{2}, 2, \frac{7}{2}, \ldots\), the common difference \(d\) can be calculated as:

• Second term minus the first term: \(2 - \frac{1}{2}\)
• Simplifying gives \(\frac{3}{2}\)

This means each subsequent term is consistently spaced \(\frac{3}{2}\) units apart.

Nth Term Formula

The nth term formula is like a treasure map in arithmetic sequences. It helps you find any term in the sequence without having to list all the previous ones. The formula is expressed as:

• \(a_n = a_1 + (n-1) \cdot d\)

Where:

• \(a_n\) is the term you're targeting (like finding the 'x' on a map)
• \(a_1\) is the starting point, or the first term of the sequence
• \(n\) is the position of the term you're interested in
• \(d\) is the common difference

In the previous example, if you want the sequence to equal 8, plug the known values \(a_1 = \frac{1}{2}\) and \(d = \frac{3}{2}\) into the equation and solve for \(n\). This formula helps confirm your path in the sequence without needing to count each step manually.

Finite Sequence

A finite sequence is like a story with a clear beginning and end. It has a specific number of terms. When you're asked to find the number of terms, you're figuring out the length of this arithmetic "story." A finite arithmetic sequence is bound by a starting and terminal point, like in our exercise where the sequence started with \(\frac{1}{2}\) and ended at 8.

• The sequence won't go on indefinitely; it has a stopping point
• Once finding the -nth term that equals the lastly defined number, you determine how long the sequence is

Identifying how many terms exist helps you understand the full scope of your arithmetic sequence and is crucial for solving related problems.

Terms in a Sequence

Every term in a sequence holds significance. They are like notes in a melody. The arrangement and the distance between them form the pattern of the sequence. In arithmetic sequences, terms advance by adding the common difference to the previous term. Here’s how it works:

• Start with the first term (given as \(a_1 = \frac{1}{2}\) in the example)
• Add the common difference to get the subsequent terms (\(2\), \(\frac{7}{2}\), etc.)

This stepwise advancement continues until the sequence's predefined end is reached (for example, the term 8 in this case). Each term is systematically derived from its predecessor and holds its position based on the sequence's rules.