Problem 64 - Algebra 2

Question

Use the given information to write an equation that represents the nth number in each arithmetic sequence. The 18th term of a sequence is 367. The 30th term of the sequence is 499. How many terms of this sequence are less than 1000?

Step-by-Step Solution

Answer

75 terms are less than 1000.

Step 1: Identify Given Information

We are given that the 18th term, \( a_{18} = 367 \), and the 30th term, \( a_{30} = 499 \), in the arithmetic sequence.

Step 2: Find Common Difference

The formula for the nth term of an arithmetic sequence is \( a_n = a_1 + (n-1) \cdot d \). For two terms in the sequence, the formula \( a_m = a_n + (m-n) \cdot d \) is used to find the common difference \( d \). Using the given terms: \( 499 = 367 + (30-18) \cdot d \), solve for \( d \):\[ 499 - 367 = 12d \]\[ 132 = 12d \]\[ d = \frac{132}{12} = 11 \]

Step 3: Write Equation for nth Term

Use one of the given terms to find the first term, \( a_1 \), by rearranging the nth term formula. Start with the 18th term representation: \[ a_{18} = a_1 + 17d = 367 \]Substitute \( d = 11 \):\[ a_1 + 17 \cdot 11 = 367 \]\[ a_1 + 187 = 367 \]\[ a_1 = 367 - 187 = 180 \]Thus, the nth term is:\[ a_n = 180 + (n-1) \cdot 11 \]

Step 4: Determine Maximum n for Terms Less Than 1000

Set the equation for the nth term less than 1000:\[ 180 + (n-1) \cdot 11 < 1000 \]Solve for \( n \):\[ 180 + 11n - 11 < 1000 \]\[ 11n < 1000 - 180 + 11 \]\[ 11n < 831 \]\[ n < \frac{831}{11} \approx 75.545 \]Since \( n \) must be an integer, \( n \leq 75 \).

Key Concepts

Common Difference

In arithmetic sequences, the term "common difference" refers to the constant difference between successive terms. This difference is what makes the sequence "arithmetic." To find it, you simply subtract any given term from the one that follows it. For example, if you know two specific terms in the sequence, like the 18th term (\(a_{18} = 367\)) and the 30th term (\(a_{30} = 499\)), you can determine the common difference, denoted as "\(d\)." The formula used is:

• \(499 = 367 + (30-18) \, d\)
• \(132 = 12d\)
• So, \(d = 11\)

This means each term in the sequence increases by 11. This consistent step helps us predict future terms easily.

Nth Term of a Sequence

The nth term of an arithmetic sequence can be calculated using a formula that helps find any term's value without listing all terms. This formula is: \(a_n = a_1 + (n-1) \, d\) where \(a_1\) is the first term, \(n\) is the term number, and \(d\) is the common difference. In the sequence we discussed, \(a_1 = 180\), and \(d = 11\). Therefore, we can write the formula for any nth term as:

• \(a_n = 180 + (n-1) \, 11\)

This formula allows rapid calculation of any term in the sequence by simply changing \(n\) without recalculating from the first term.

Sequence Equation

The sequence equation is a mathematical representation that provides the value of any term in the sequence based on its position. For arithmetic sequences, it takes the form of:\(a_n = a_1 + (n-1) \, d\).This equation is essential because it embodies the pattern of the sequence, telling us how to derive every single term by knowing where you begin and the consistent step you take. Returning to our example, by plugging the initial values \(a_1 = 180\) and \(d = 11\) into the nth term formula, we derive:

• \(a_n = 180 + (n-1) \, 11\).

This equation precisely formulates the entire sequence and allows us to calculate any term in the series efficiently.

Arithmetic Progression

An arithmetic progression is a series of numbers in which the difference between consecutive terms remains constant. This constant is known as the "common difference." Such sequences are straightforward and predictable, as demonstrated in our example. The clear pattern in an arithmetic progression lets you determine relationships between terms easily. For instance, using terms such as the 18th and 30th to find critical details like the common difference or even to derive the complete sequence gives you a practical approach to understanding the flow of numbers.

• Each term can be realized by adding the common difference to the preceding term, allowing this methodical increase to define the entire sequence.
• It is useful for solving problems involving maximum or specific terms in a scenario, as the order and repetition allow for precise calculations.

Understanding arithmetic progression is crucial for both understanding fundamental mathematical concepts and for their application in real-world mathematical scenarios.