Problem 8 - Precalculus Mathematics for Calculus

Question

A sequence is given. (a) Find the first five terms of the sequence. (b) What is the common difference \(d ?\) (c) Graph the terms you found in (a). $$a_{n}=\frac{1}{2}(n-1)$$

Step-by-Step Solution

Answer

(a) 0, 0.5, 1, 1.5, 2; (b) \(d = 0.5\); (c) Points are collinear.

Step 1: Calculate the First Term

To find the first term of the sequence, substitute \( n = 1 \) into the formula. Thus, \( a_1 = \frac{1}{2}(1-1) = \frac{1}{2}(0) = 0 \).

Step 2: Calculate the Second Term

Substitute \( n = 2 \) to find the second term: \( a_2 = \frac{1}{2}(2-1) = \frac{1}{2}(1) = \frac{1}{2} \).

Step 3: Calculate the Third Term

Substitute \( n = 3 \) to find the third term: \( a_3 = \frac{1}{2}(3-1) = \frac{1}{2}(2) = 1 \).

Step 4: Calculate the Fourth Term

Substitute \( n = 4 \) to find the fourth term: \( a_4 = \frac{1}{2}(4-1) = \frac{1}{2}(3) = \frac{3}{2} \).

Step 5: Calculate the Fifth Term

Substitute \( n = 5 \) to find the fifth term: \( a_5 = \frac{1}{2}(5-1) = \frac{1}{2}(4) = 2 \).

Step 6: Identify the Common Difference

To find the common difference \( d \), subtract consecutive terms: \( a_2 - a_1 = \frac{1}{2} - 0 = \frac{1}{2} \), \( a_3 - a_2 = 1 - \frac{1}{2} = \frac{1}{2} \), and so on. Therefore, the common difference \( d \) is \( \frac{1}{2} \).

Step 7: Graph the Terms

Plot the points calculated in steps 1 through 5 on a graph. The x-axis represents \( n \) and the y-axis represents \( a_n \). You should plot the points: (1, 0), (2, \( \frac{1}{2} \)), (3, 1), (4, \( \frac{3}{2} \)), (5, 2). Since this is an arithmetic sequence, the points should form a straight line.

Key Concepts

Common Difference

The common difference in an arithmetic sequence is pivotal because it defines the consistent incremental pattern between terms. In the sequence given by the formula \(a_{n}=\frac{1}{2}(n-1)\), you can observe a steady increase as each succeeding term is calculated. To determine this difference, subtract any two consecutive terms. For example, to find the common difference \(d\), we can take the first two terms of the sequence: \(a_1 = 0\) and \(a_2 = \frac{1}{2}\). Subtract \(a_1\) from \(a_2\):

• \(d = a_2 - a_1 = \frac{1}{2} - 0 = \frac{1}{2}\)

With all terms consistently increasing by \(\frac{1}{2}\), we confirm that our sequence follows this common difference. This difference demonstrates the linear relationship inherent in arithmetic sequences.

Graphing Sequences

Graphing an arithmetic sequence helps visualize how the series progresses over time. Each term of the sequence corresponds to a point on the graph. For our sequence \(a_{n}=\frac{1}{2}(n-1)\), we calculated the first five terms: \(0, \frac{1}{2}, 1, \frac{3}{2}, 2\). These terms translate into the following points on the graph:

• (1, 0)
• (2, \(\frac{1}{2}\))
• (3, 1)
• (4, \(\frac{3}{2}\))
• (5, 2)

When plotted, you'll notice these points form a straight line. This linearity is characteristic of arithmetic sequences, indicating a constant rate of increase (or decrease) across terms. On a graph, the x-axis typically represents the term number \(n\) and the y-axis represents the value \(a_n\). Connecting these points with a straight line showcases the regularity of the sequence's growth.

Precalculus Mathematics

Understanding arithmetic sequences is a foundational skill in precalculus mathematics. It involves exploring sequences, a type of mathematical progression that frequently appears across various mathematical concepts. In the context of the sequence given by \(a_{n}=\frac{1}{2}(n-1)\), the goal is to learn not just about the sequence itself, but to also comprehend its broader applications and implications:

• Sequences like this are useful in modeling situations where something accumulates at a constant rate, such as savings accounts or uniform motion scenarios.
• The arithmetic sequence serves as a building block for understanding more complex mathematical topics like series and calculus.
• Grasping such patterns assists students in developing logical thinking and enhancing problem-solving skills.

Through mastering such sequences, students prepare themselves for further mathematical studies, where recognizing patterns and predicting outcomes form core skills.