Problem 32 - Algebra and Trigonometry

Question

\(27-36\) . Determine the common difference, the fifth term, the nth term, and the 100 th term of the arithmetic sequence. $$ \frac{7}{6}, \frac{5}{3}, \frac{13}{6}, \frac{8}{3}, \dots $$

Step-by-Step Solution

Answer

Common difference: \( \frac{1}{2} \); Fifth term: \( \frac{17}{6} \); nth term: \( a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} \); 100th term: \( \frac{152}{3} \).

Step 1: Identify the first term

The first term of the sequence is the first number given, which is \( \frac{7}{6} \).

Step 2: Find the common difference

The common difference in an arithmetic sequence is obtained by subtracting the first term from the second term. So, we calculate: \( d = \frac{5}{3} - \frac{7}{6} \). To perform the subtraction, convert \( \frac{5}{3} \) to \( \frac{10}{6} \). Then, calculate \( d = \frac{10}{6} - \frac{7}{6} = \frac{3}{6} = \frac{1}{2} \). Thus, the common difference is \( \frac{1}{2} \).

Step 3: Calculate the fifth term

We know that each term in the sequence increases by the common difference, \( \frac{1}{2} \). Starting with the first term \( \frac{7}{6} \), we find the fifth term by adding \( \frac{1}{2} \) four times.\[\begin{align*}2nd \ term &: \frac{7}{6} + \frac{1}{2} = \frac{5}{3} \3rd \ term &: \frac{5}{3} + \frac{1}{2} = \frac{13}{6} \4th \ term &: \frac{13}{6} + \frac{1}{2} = \frac{8}{3} \5th \ term &: \frac{8}{3} + \frac{1}{2} = \frac{17}{6}\end{align*}\]The fifth term is \( \frac{17}{6} \).

Step 4: Determine the nth term formula

The formula for the nth term \( a_n \) of an arithmetic sequence is given by \( a_n = a_1 + (n-1)d \), where \( a_1 \) is the first term and \( d \) is the common difference. For this sequence, \( a_1 = \frac{7}{6} \) and \( d = \frac{1}{2} \). Thus, the formula becomes: \[ a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} \].

Step 5: Calculate the 100th term

Substitute \( n = 100 \) into the nth term formula:\[a_{100} = \frac{7}{6} + (100 - 1) \cdot \frac{1}{2}\]Calculate:\[= \frac{7}{6} + 99 \cdot \frac{1}{2} \]Convert \( 99 \cdot \frac{1}{2} \) to fractions: \( 99 \cdot \frac{1}{2} = \frac{99}{2} \).Combine the fractions:\[a_{100} = \frac{7}{6} + \frac{297}{6} = \frac{304}{6} = \frac{152}{3}\]So, the 100th term is \( \frac{152}{3} \).

Key Concepts

Common Difference

The common difference in an arithmetic sequence is crucial for understanding how the sequence progresses from one term to the next. It is consistent throughout the sequence, meaning that each term increases or decreases by the same amount.
To find the common difference, you subtract the first term from the second term. In this example, the sequence begins with \( \frac{7}{6}, \frac{5}{3}, \frac{13}{6}, \dots \). You take the second term \( \frac{5}{3} \) and subtract the first term \( \frac{7}{6} \):

• First, convert \( \frac{5}{3} \) to an equivalent fraction with the same denominator as \( \frac{7}{6} \), which is \( \frac{10}{6} \).
• Subtract \( \frac{7}{6} \) from \( \frac{10}{6} \) to get the common difference \( \frac{3}{6} = \frac{1}{2} \).

With the common difference, you gain insight into how the sequence behaves and can predict any term's position.

Nth Term Formula

The nth term formula is what allows you to find any term in an arithmetic sequence without having to calculate all preceding terms. This saves time and effort, especially for large sequences. The formula is expressed as:

• \( a_n = a_1 + (n-1)d \)

where:

• \( a_1 \) is the first term of the sequence.
• \( d \) stands for the common difference.
• \( n \) represents the term number you want to find.

For the sequence \( \frac{7}{6}, \frac{5}{3}, \frac{13}{6}, \dots \), we have \( a_1 = \frac{7}{6} \) and \( d = \frac{1}{2} \). Substituting these values into the formula provides:

• \( a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} \)

This formula can now be used to find any term in this arithmetic sequence with ease.

Fifth Term

Finding specific terms like the fifth term in an arithmetic sequence helps solidify your understanding of the sequence. Knowing the common difference, you can incrementally calculate each subsequent term. Starting with the first term \( \frac{7}{6} \), and common difference \( \frac{1}{2} \):

• Second term: \( \frac{7}{6} + \frac{1}{2} = \frac{5}{3} \)
• Third term: \( \frac{5}{3} + \frac{1}{2} = \frac{13}{6} \)
• Fourth term: \( \frac{13}{6} + \frac{1}{2} = \frac{8}{3} \)
• Fifth term: \( \frac{8}{3} + \frac{1}{2} = \frac{17}{6} \)

As shown, the fifth term of this sequence is \( \frac{17}{6} \). This step-by-step calculation showcases how consecutive terms are built upon each other using the common difference.

100th Term

Calculating a term so far along in the sequence, like the 100th term, exemplifies the power of the nth term formula. Instead of computing each term until the 100th, use the formula to jump straight to the desired term number.
First, recall the nth term formula:

• \( a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} \)

Set \( n = 100 \) for the 100th term:

• \( a_{100} = \frac{7}{6} + (100 - 1) \cdot \frac{1}{2} \)
• Calculate \( 99 \cdot \frac{1}{2} = \frac{99}{2} \).
• Combine the fractions: \( a_{100} = \frac{7}{6} + \frac{297}{6} = \frac{304}{6} \).
• Simplify to find \( \frac{152}{3} \).

This shows the remarkable ability of the formula to swiftly and accurately yield any term, including the 100th, in the sequence.