Problem 49 - Algebra 2

Question

Find the first five terms of each arithmetic sequence described. \(a_{1}=\frac{4}{3}, d=-\frac{1}{3}\)

Step-by-Step Solution

Answer

The first five terms are \(\frac{4}{3}, 1, \frac{2}{3}, \frac{1}{3}, 0\).

Step 1: Recall the formula for an arithmetic sequence

An arithmetic sequence is defined by its first term, \(a_1\), and a common difference, \(d\). The \(n\)-th term of an arithmetic sequence can be found using the formula: \(a_n = a_1 + (n-1) imes d\).

Step 2: Find the second term

The second term, \(a_2\), is calculated by substituting \(n=2\) into the arithmetic sequence formula. Thus: \(a_2 = \frac{4}{3} + (2-1) \times -\frac{1}{3}\). Simplifying, \(a_2 = \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1\).

Step 3: Find the third term

The third term, \(a_3\), is calculated using \(n=3\) in the formula. Thus: \(a_3 = \frac{4}{3} + (3-1) \times -\frac{1}{3}\). Simplifying, \(a_3 = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}\).

Step 4: Find the fourth term

To find the fourth term, \(a_4\), use \(n=4\) in the formula: \(a_4 = \frac{4}{3} + (4-1) \times -\frac{1}{3}\). Simplifying, \(a_4 = \frac{4}{3} - 1 = \frac{1}{3}\).

Step 5: Find the fifth term

The fifth term, \(a_5\), is found by substituting \(n=5\) into the formula: \(a_5 = \frac{4}{3} + (5-1) \times -\frac{1}{3}\). Simplifying, \(a_5 = \frac{4}{3} - \frac{4}{3} = 0\).

Step 6: List the first five terms

After computing each term, the first five terms of the sequence are: \(\frac{4}{3}, 1, \frac{2}{3}, \frac{1}{3}, 0\).

Key Concepts

Common Difference

In an arithmetic sequence, the term "common difference" refers to the constant amount added to each term to get the next term. This value is represented as \(d\). It is consistent throughout the entire sequence, meaning that every term is separated by this same difference. Knowing the common difference is crucial because it helps in predicting future terms. In the given problem, the common difference is \(-\frac{1}{3}\). This implies that each term is \(\frac{1}{3}\) less than the previous one.
For example:

• Starting with the first term \(\frac{4}{3}\), the second term becomes \(\frac{4}{3} - \frac{1}{3} = 1\).
• The third term, \(1 - \frac{1}{3} = \frac{2}{3}\), continues this pattern.

The common difference allows us to write a general formula to find any term in the sequence, which is a powerful tool for solving various mathematical problems.

Sequence Formula

The sequence formula in arithmetic sequences is a mathematical expression used to find any specific term in the sequence. This is represented by the expression: \(a_n = a_1 + (n-1) \times d\).
Here:

• \(a_n\) is the \(n\)-th term you want to find.
• \(a_1\) is the first term, or starting point, of the sequence.
• \(d\) is the common difference.
• \(n\) is the term position in the sequence (e.g., 1st, 2nd, 3rd...)

This formula makes it easy to calculate any term without having to list all previous terms. For instance, to find the fifth term in our sequence, substitute \(a_1 = \frac{4}{3}\), \(d = -\frac{1}{3}\), and \(n = 5\) into the formula:

\(a_5 = \frac{4}{3} + (5-1) \times (-\frac{1}{3})\)
\(= \frac{4}{3} - \frac{4}{3} = 0\)
The formula shows how each piece of information contributes to identifying any particular sequence term, while also simplifying the calculation process.

First Term

In an arithmetic sequence, the first term, often denoted as \(a_1\), is the initial starting point. It is the base value from which all subsequent terms are derived by adding the common difference. The first term is essential because it sets the stage for the entire sequence.
In the problem we're examining, the first term given is \(\frac{4}{3}\). From here, using the common difference, every subsequent term is calculated.

• To find the second term, add the common difference to the first term: \(\frac{4}{3} - \frac{1}{3} = 1\).
• The process continues similarly for each following term, always starting calculations from the first term.

Thus, the first term is like the "seed" that starts growing the sequence, from which every future term is cultivated through the application of the common difference and the sequence formula.