Problem 31 - Precalculus: Mathematics for Calculus

Question

Find the \(n\)th term of a sequence whose first several terms are given. \(-2,3,8,13, \dots\)

Step-by-Step Solution

Answer

The \(n\)th term of the sequence is \(5n - 7\).

Step 1: Identify the Pattern

Observe the given sequence: \(-2, 3, 8, 13, \dots\). Calculate the difference between consecutive terms to see if there is a common difference. \(3 - (-2) = 5\), \(8 - 3 = 5\), and \(13 - 8 = 5\). The common difference is 5, suggesting it's an arithmetic sequence.

Step 2: Recall the Formula for Arithmetic Sequences

The formula to find the \(n\)th term of an arithmetic sequence is: \(a_n = a + (n-1) \times d\), where \(a\) is the first term and \(d\) is the common difference.

Step 3: Substitute Known Values

In the formula \(a_n = a + (n-1) \times d\), substitute \(a = -2\) (the first term) and \(d = 5\) (the common difference). The formula becomes: \(a_n = -2 + (n-1) \times 5\).

Step 4: Simplify the Expression

Distribute and simplify the expression from the previous step: \(a_n = -2 + 5n - 5 = 5n - 7\).

Step 5: Write the General Term

The \(n\)th term of the sequence can be represented as \(a_n = 5n - 7\).

Key Concepts

nth term formula

In any arithmetic sequence, we often want to find a specific term, known as the \(n\)th term. Luckily, there's a formula for this! It's given by \[ a_n = a + (n-1) \times d \], where \(a\) is the first term of the sequence and \(d\) is the common difference. This formula helps us find any term by plugging in the position number of that term (\(n\)).

• \(a\) is the initial value where our sequence starts.
• \(n\) minus 1 represents the number of jumps we've made in the sequence.
• \(d\) is the uniform amount added to each term to get the next one.

Once we replace these symbols with their known values, we can calculate any term of the sequence without listing all the terms up to that point.

common difference

A key component of an arithmetic sequence is its common difference. The common difference, \(d\), is the consistent amount added to each term to arrive at the next term. For example, in our sequence \(-2, 3, 8, 13, \ldots \), the difference is 5.

• To find the common difference, subtract any term from the term that follows it.
• In our example, \(3 - (-2) = 5\), and the same value is found between each consecutive pair.

Having a common difference is what classifies this sequence as arithmetic. Once identified, it is used in the \(n\)th term formula to find the value of further terms in the sequence.

sequence pattern

Recognizing and understanding the pattern is crucial in arithmetic sequences, as they follow a predictable pattern thanks to their constant difference. The sequence pattern becomes evident as we identify the common difference.

• Each term is a simple addition of the common difference to the previous term.
• In our example: starting from \(-2\), we add 5 to get 3, again 5 to get 8, and this keeps going.

Observing this pattern helps to verify our calculations are correct and ensures the sequence remains on track. Patterns also allow us to easily grasp how sequences extend beyond the given terms.

precalculus

While arithmetic sequences themselves might seem straightforward, they are fundamental to the study of precalculus. They introduce concepts that are foundational to more complex mathematical ideas.

• Understanding sequences paves the way to learning about series and summations.
• Sequences give students a glimpse into the world of functions, since they can be described using equations.

Applying these concepts in precalculus builds a bridge to calculus, where the study of changes and patterns continues on a broader scale. Mastering these sequences is a stepping stone to solving real-world problems using mathematics.