Problem 11 - College Algebra

Question

For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference. $$ a_{1}=0, d=\frac{2}{3} $$

Step-by-Step Solution

Answer

The first five terms are: 0, \(\frac{2}{3}\), \(\frac{4}{3}\), 2, \(\frac{8}{3}\).

Step 1: Understanding the Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference, denoted by \( d \).

Step 2: Identify the First Term

The first term of the arithmetic sequence is given as \( a_1 = 0 \). This will be the starting point for generating the sequence.

Step 3: Common Difference

The common difference \( d \) is provided as \( \frac{2}{3} \). This means that each term is obtained by adding \( \frac{2}{3} \) to the previous term.

Step 4: Calculate the Second Term

Use the formula for the second term: \( a_2 = a_1 + d \). Plug in the values: \( a_2 = 0 + \frac{2}{3} = \frac{2}{3} \).

Step 5: Calculate the Third Term

Use the formula for the third term: \( a_3 = a_2 + d \). Substitute the known values: \( a_3 = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \).

Step 6: Calculate the Fourth Term

Use the formula for the fourth term: \( a_4 = a_3 + d \). Substitute: \( a_4 = \frac{4}{3} + \frac{2}{3} = 2 \).

Step 7: Calculate the Fifth Term

Use the formula for the fifth term: \( a_5 = a_4 + d \). Substitute: \( a_5 = 2 + \frac{2}{3} = \frac{8}{3} \).

Step 8: Write the First Five Terms

The first five terms of the sequence are: \( 0, \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \).

Key Concepts

Understanding the Common Difference in Arithmetic Sequences

When working with arithmetic sequences, a key aspect to grasp is the 'common difference'. This is a fixed, constant value that you add to each term in the sequence to get to the next one. You can think of it as the "step" between numbers.
For example, in the problem where the first term is given as 0 and the common difference is \( \frac{2}{3} \), this means each step from one number to the next is always \( \frac{2}{3} \).

Here’s how you find the common difference:

• Take any two consecutive terms in the sequence.
• Subtract the first term from the second term.

This subtraction will always give you the common difference if the sequence is arithmetic.
So, understanding this concept helps to predict any term in the sequence, making arithmetic sequences a straightforward yet powerful mathematical tool.

Using Sequence Formulas in Arithmetic Sequences

In arithmetic sequences, a core component is the use of sequence formulas. These are mathematical expressions used to calculate any term in the sequence.

The formula to find the nth term of an arithmetic sequence is:
\[ a_n = a_1 + (n-1) \cdot d \]

In this formula:

• \( a_n \) is the nth term you're trying to find.
• \( a_1 \) is the first term of the sequence.
• \( n \) is the term number.
• \( d \) is the common difference.

By substituting the values for these variables, you can find any term's value.
Consider the problem's sequence where \( a_1 = 0 \) and \( d = \frac{2}{3} \). Using the formula:\( a_2 = a_1 + d = 0 + \frac{2}{3} = \frac{2}{3} \).
The formula ensures consistency and accuracy in calculations, especially when working with longer sequences.

Term Calculation in Arithmetic Sequences

Term calculation in arithmetic sequences involves using both the understanding of common difference and sequence formulas.
Let's break it down:

Firstly, begin with the first term, as this is your starting point.

• If the first term, \( a_1 \), is given as 0, this will be your foundation.
• Then add the common difference to find sequential terms.

For instance, to find the second term \( a_2 \): use
\( a_2 = a_1 + d = 0 + \frac{2}{3} = \frac{2}{3} \).

Then, to find the third term \( a_3 \):
\( a_3 = a_2 + d = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \).

Hence, this logical process continues by calculating term after term.
This technique assures you calculate each subsequent term accurately while always knowing the difference between them.

With practice, calculating terms becomes intuitive, and you can effortlessly determine as many terms as needed.