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\(5-8=\) 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). $$ a_{n}=\frac{1}{2}(n-1) $$

Short Answer

Expert verified
The first five terms are 0, \(\frac{1}{2}\), 1, \(\frac{3}{2}\), 2, and the common difference \(d\) is \(\frac{1}{2}\).

Step by step solution

01

Understand the Sequence Formula

The given sequence is defined as \(a_n = \frac{1}{2}(n-1)\). This means that the term \(a_n\) depends linearly on the index \(n\). Each term is derived by substituting \(n\) with its position in the sequence.
02

Calculate the First Term

To find the first term \(a_1\), substitute \(n = 1\) into the formula: \[a_1 = \frac{1}{2}(1-1) = 0\].So, the first term is \(0\).
03

Calculate the Second Term

For the second term \(a_2\), substitute \(n = 2\): \[a_2 = \frac{1}{2}(2-1) = \frac{1}{2}\].Therefore, the second term is \(\frac{1}{2}\).
04

Calculate the Third Term

To find the third term \(a_3\), substitute \(n = 3\): \[a_3 = \frac{1}{2}(3-1) = 1\].Thus, the third term is \(1\).
05

Calculate the Fourth Term

For the fourth term \(a_4\), substitute \(n = 4\): \[a_4 = \frac{1}{2}(4-1) = \frac{3}{2}\].The fourth term is \(\frac{3}{2}\).
06

Calculate the Fifth Term

To find the fifth term \(a_5\), substitute \(n = 5\): \[a_5 = \frac{1}{2}(5-1) = 2\].As a result, the fifth term is \(2\).
07

Determine the Common Difference

The common difference \(d\) in an arithmetic sequence is the difference between consecutive terms. With the sequence: \(0, \frac{1}{2}, 1, \frac{3}{2}, 2\), we calculate:\[\frac{1}{2} - 0 = \frac{1}{2}, \quad 1 - \frac{1}{2} = \frac{1}{2}, \quad \frac{3}{2} - 1 = \frac{1}{2}, \quad 2 - \frac{3}{2} = \frac{1}{2}\].Thus, \(d = \frac{1}{2}\).
08

Graph the Sequence Terms

Plot the points \((1,0), (2,\frac{1}{2}), (3,1), (4,\frac{3}{2}), (5,2)\) on the Cartesian plane. These points should form a straight line, confirming the linear nature of the sequence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Difference
In an arithmetic sequence, a common difference is a constant value that you add (or subtract) to each term to get the next term in the sequence. Understanding the common difference helps identify the pattern of the sequence.
In the given sequence, each term increases by the same amount from the previous term. This sequence is a linear sequence because each term progresses by adding a **common difference**. Let's break down the calculation of the common difference using the sequence from our exercise.
  • Start with the first few terms provided: 0, \( \frac{1}{2} \), 1, \( \frac{3}{2} \), and 2.
  • Calculate the difference between consecutive terms. For example, subtract the first term from the second: \( \frac{1}{2} - 0 = \frac{1}{2} \).
  • Repeat the process for all consecutive terms: \( 1 - \frac{1}{2} = \frac{1}{2} \), \( \frac{3}{2} - 1 = \frac{1}{2} \), and \( 2 - \frac{3}{2} = \frac{1}{2} \).
The difference is consistently \( \frac{1}{2} \). This consistent value is called the **common difference** and confirms the arithmetic nature of the sequence.
Recognizing this uniform pattern is key in understanding arithmetic sequences.
Linear Sequences
A **linear sequence** is a series of numbers in which the difference between consecutive numbers is constant. It’s called "linear" because if you graph the terms of the sequence, they form a straight line. This property arises due to the common difference, as already demonstrated.
In our exercise, the given sequence was defined by the formula: \( a_n = \frac{1}{2}(n - 1) \), which is a linear function of \( n \). Here's how linear sequences operate:
  • **Substitute values of \( n \):** To find sequence terms, substitute increasing natural numbers for \( n \) into the formula.
  • **Uniform increase:** Each term is increased by the same amount (the common difference) from the previous term as \( n \) increases.
  • **Predictability:** Because the rule is consistent, linear sequences are predictable. You can easily find any term in the sequence by continuing to apply the formula.
Linear sequences provide a straightforward way to model simple incremental changes and are commonly used in algebra to study progression and predict future terms.
Graphing Sequences
Graphing a sequence helps visualize the pattern and analyze the sequence’s nature. When you graph an arithmetic sequence, you will always see a pattern.
Let's see how it looks using the terms from the exercise's sequence.
  • To start graphing, prepare the Cartesian plane with an appropriate scale that includes all the sequence terms.
  • Next, mark each term as a point on the plane where the x-coordinate represents the term's position in the sequence, and the y-coordinate represents the value of the term.
  • For the sequence \( (1,0), (2,\frac{1}{2}), (3,1), (4,\frac{3}{2}), (5,2) \):
The points align in a straight line which indicates that the sequence is linear. Each step along the x-axis (from one term to the next) represents a uniform movement along the y-axis by the common difference of \( \frac{1}{2} \).
This visual approach confirms the theoretical calculation of the common difference and demonstrates the sequence's linear growth. Graphing is a powerful tool to verify and understand sequences, making abstract patterns concrete and easier to explore.

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