Problem 52 - College Algebra

Question

For the following exercises, write an explicit formula for each arithmetic sequence. $$ a=\left\\{-5,-\frac{10}{3},-\frac{5}{3}, \ldots\right\\} $$

Step-by-Step Solution

Answer

\(a_n = \frac{5}{3}n - \frac{25}{3}\)

Step 1: Identify the first term

The first term of the sequence is the first element in the list provided. In this sequence, the first term is \(-5\).

Step 2: Determine the common difference

To find the common difference \(d\) in an arithmetic sequence, subtract the first term from the second term. Here, \(-\frac{10}{3} - (-5) \) gives a common difference of \(\frac{5}{3}\).

Step 3: Write the explicit formula

The explicit formula for an arithmetic sequence can be written as \(a_n = a_1 + (n-1) \cdot d\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number. Substituting \(a_1 = -5\) and \(d = \frac{5}{3}\), the explicit formula becomes: \(a_n = -5 + (n-1) \cdot \frac{5}{3}\).

Step 4: Simplify the explicit formula

Distribute the common difference \(\frac{5}{3}\) through \((n-1)\). This gives \(a_n = -5 + \frac{5}{3}n - \frac{5}{3}\). Simplify the constant terms to get \(a_n = \frac{5}{3}n - \frac{10}{3} - 5\). Further simplify to \(a_n = \frac{5}{3}n - \frac{25}{3}\).

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a set of numbers where the difference between consecutive terms remains constant. This constant difference is known in algebra as the "common difference." In simple terms, each number in the sequence is obtained by adding the common difference to the previous number. This systematic addition creates a predictable and linear progression of numbers.
For example, consider the arithmetic sequence:

• 2, 5, 8, 11,...

In this sequence, each number increases by 3, which is the common difference. Arithmetic sequences are foundational in understanding mathematical patterns and are exploitable in various real-life scenarios, like calculating monthly savings or project timelines.

Common Difference

The common difference (\(d\)) is a crucial part of an arithmetic sequence. This is the consistent difference between any two consecutive terms in the sequence.
To find the common difference, simply subtract the first term from the second term. In general terms, if the sequence is \(a_1, a_2, a_3, \ldots\), the common difference \(d = a_2 - a_1\).
For example, in the sequence \(-5, -\frac{10}{3}, -\frac{5}{3}, \ldots\), the common difference is \(\frac{5}{3}\) because: \(-\frac{10}{3} - (-5) = \frac{5}{3}\).
Understanding the common difference is vital because it determines the nature and direction (increasing or decreasing) of the sequence.

Explicit Formula

In mathematics, an explicit formula provides a direct means of finding any term in a sequence without needing the previous term. For arithmetic sequences, the explicit formula is:

• \(a_n = a_1 + (n-1)\cdot d\)

Where:

• \(a_n\) is the term you want to find
• \(a_1\) is the first term in the sequence
• \(d\) is the common difference
• \(n\) is the term number

This formula allows for fast calculations, especially when dealing with large term numbers, because you do not need to sequentially add the common difference to finding a specific term. Given our sequence from the exercise, the explicit formula is \(a_n = \frac{5}{3}n - \frac{25}{3}\), which efficiently finds any term directly.

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. It serves as a unifying thread of understanding various mathematical problems and ideas, like solving for unknowns, symbolized typically by letters such as \(x\) and \(n\).
Algebra supports concepts like arithmetic sequences by providing these symbolic expressions and formulas that model and solve real-world problems. For instance, using the explicit formula in arithmetic sequences, we can apply algebraic principles to simplify expressions and find any term in the sequence. In the given exercise, algebra helps us rearrange and simplify: \(a_n = -5 + \frac{5}{3}n - \frac{5}{3}\) to \(a_n = \frac{5}{3}n - \frac{25}{3}\) ensuring effective solutions.