Problem 40 - Precalculus: Mathematics for Calculus

Question

Determine the common difference, the fifth term, the \(n\) th 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

The common difference is \(\frac{1}{2}\), the fifth term is \(\frac{19}{6}\), the nth term is \(a_n = \frac{7+3n}{6}\), and the 100th term is \(\frac{307}{6}\).

Step 1: Identify the Common Difference

An arithmetic sequence increases by a constant difference from one term to the next. Calculate the difference between the second term \(\frac{5}{3}\) and the first term \(\frac{7}{6}\):\[\text{Common difference } d = \frac{5}{3} - \frac{7}{6} = \frac{10}{6} - \frac{7}{6} = \frac{3}{6} = \frac{1}{2}.\]

Step 2: Calculate the Fifth Term

Use the first term and the common difference to find the fifth term of the sequence.The general formula for the \(n\)th term in an arithmetic sequence is\[ a_n = a_1 + (n-1) \, d. \]Substitute the values \(a_1 = \frac{7}{6}\), \(d = \frac{1}{2}\), and \(n = 5\):\[a_5 = \frac{7}{6} + (5-1) \cdot \frac{1}{2} = \frac{7}{6} + 2 = \frac{7}{6} + \frac{12}{6} = \frac{19}{6}.\]

Step 3: Derive the General Formula for the nth Term

Use the known formula for the \(n\)th term:\[ a_n = a_1 + (n-1) \, d. \]Substitute the known values \(a_1 = \frac{7}{6}\) and \(d = \frac{1}{2}\):\[ a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} = \frac{7}{6} + \frac{n-1}{2}. \]To simplify, convert \(\frac{7}{6}\) to have a common denominator of 2:\[ \frac{7}{6} = \frac{7}{6} = \frac{7}{6}. \] This gives us:\[ a_n = \frac{7}{6} + \frac{n}{2} - \frac{1}{2}. \] Simplify to get:\[ a_n = \frac{7+3n}{6}. \]

Step 4: Calculate the 100th Term

Substitute \(n = 100\) in the derived \(n\)th term equation:\[ a_{100} = \frac{7+3(100)}{6} = \frac{7+300}{6} = \frac{307}{6}. \]

Key Concepts

Common difference

In an arithmetic sequence, a crucial element is the common difference, denoted as \( d \). This is the constant difference between consecutive terms. Let's say you have two successive numbers in a sequence: \( a_2 \) and \( a_1 \). The common difference can be found using the formula:

• \( d = a_2 - a_1 \)

Applying this to our sequence,

• First term \( a_1 = \frac{7}{6} \)
• Second term \( a_2 = \frac{5}{3} \)
• Calculate: \( d = \frac{5}{3} - \frac{7}{6} \)
• Convert \( \frac{5}{3} \) to have a common denominator with \( \frac{7}{6} \): it becomes \( \frac{10}{6} \)
• Result: \( d = \frac{10}{6} - \frac{7}{6} = \frac{3}{6} = \frac{1}{2} \)

The common difference tells us how much we add to get the next term in the sequence. This consistent interval makes arithmetic sequences straightforward and predictable.

Nth term formula

The nth term formula of an arithmetic sequence allows you to find any term in the sequence without calculating each consecutive term. The formula is given by:

• \( a_n = a_1 + (n-1) \, d \)

Here,

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

By plugging in our particular sequence values:

• \( a_1 = \frac{7}{6} \)
• \( d = \frac{1}{2} \)
• \( a_n = \frac{7}{6} + (n-1) \cdot \frac{1}{2} \)

Simplify to find the formula:

• \( a_n = \frac{7}{6} + \frac{n-1}{2} = \frac{7+3n}{6} \)

This formula efficiently gives us any term's value in this sequence. For instance, using \( n = 5 \) finds the fifth term directly without computing the first four terms.

Arithmetic progression

An arithmetic progression, or arithmetic sequence is a sequence of numbers where each term after the first is obtained by adding a constant difference, \( d \), to the preceding term. This simple pattern offers predictability:

• From the formula \( a_n = a_1 + (n-1) \, d \), the entire sequence is defined once the first term and common difference are known.

The benefits are highlighted when finding any term in the series, such as the 100th term:

• \( a_{100} = \frac{7+3 \times 100}{6} = \frac{307}{6} \)

An arithmetic progression's uniformity makes calculation systematic, removing guesswork. Understanding this pattern equips you to handle tasks efficiently, like determining how far a sequence might go or locating specific terms without iterative calculations.