Problem 33 - Precalculus with Limits

Question

In Exercises 29 - 34, write the first five terms of the geometric sequence. Determine the common ratio and write the \( n \)th term of the sequence as a function of \( n \). \( a_1 = 6, a_{k + 1} = -\dfrac{3}{2} a_k \)

Step-by-Step Solution

Answer

The first five terms of the sequence are 6, -9, 13.5, -20.25, 30.375. The common ratio is -3/2, and the nth term of the sequence can be written as \( a_n = 6 \times (-\dfrac{3}{2})^{(n - 1)} \).

Step 1: Find the first five terms

Using the recursive formula and the first term, each subsequent term of the sequence can be found by multiplying the previous term by -3/2. Thus, the first five terms are as follows: \( a_1 = 6 \), \( a_2 = -\dfrac{3}{2} \times 6 = -9 \), \( a_3 = -\dfrac{3}{2} \times -9 = 13.5 \), \( a_4 = -\dfrac{3}{2} \times 13.5 = -20.25 \), \( a_5 = -\dfrac{3}{2} \times -20.25 = 30.375 \).

Step 2: Identify the common ratio

The common ratio of a geometric sequence is the factor by which each term is multiplied to get the next term. In this sequence, each term is multiplied by -3/2 to get to the next term. Therefore, the common ratio \( r \) is -3/2.

Step 3: Write the nth term as a function of n

The nth term of a geometric sequence can be found using the formula \( a_n = a_1 \times r^{(n - 1)} \). In this case, the first term \( a_1 \) is 6 and the common ratio \( r \) is -3/2, so the nth term can be written as \( a_n = 6 \times (-\dfrac{3}{2})^{(n - 1)} \).

Key Concepts

Common Ratio

In a geometric sequence, the common ratio plays an essential role, serving as the factor by which we multiply each term to reach the next. Identifying the common ratio helps us understand the pattern within the sequence. Consider the given sequence where the first term is 6. To find the second term, we multiply the first term by \(-\dfrac{3}{2}\).
This results in a second term of -9. Continuing this process, we multiply each subsequent term by \(-\dfrac{3}{2}\) to obtain the next.

• For instance, \(-\dfrac{3}{2} \times -9 = 13.5\) for the third term.
• This pattern continues, showing how the common ratio is consistently applied to obtain each following term.

The constant use of the common ratio ensures that the sequence remains geometric, maintaining a regular pattern between consecutive terms.

Nth Term

The concept of the nth term is crucial in understanding any sequence, particularly geometric sequences. When we talk about the nth term, we refer to a function that represents any term in the sequence based on its position, n.
In a geometric sequence, the nth term can be calculated using the formula \(a_n = a_1 \times r^{(n-1)}\).Here, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

• For our sequence, the first term \(a_1 = 6\).
• The common ratio is \(-\dfrac{3}{2}\).
• Therefore, the nth term becomes \(a_n = 6 \times \left(-\dfrac{3}{2}\right)^{n-1}\).

This formula allows us to quickly calculate any term in the sequence, offering insight into its behavior and pattern without calculating each individual term sequentially.

Recursive Formula

The recursive formula is a method for finding the terms of a sequence using one or more of its preceding terms. It highlights the relationship between consecutive terms. In the given exercise, the recursive formula is \(a_{k+1} = -\dfrac{3}{2} \times a_k\).
This tells us how to compute a term \(a_{k+1}\) based on the previous term \(a_k\).

• To find the second term, start with \(a_1 = 6\), and apply the formula: \(a_2 = -\dfrac{3}{2} \times 6 = -9\).
• For the third term: \(a_3 = -\dfrac{3}{2} \times (-9) = 13.5\).
• This pattern follows for all subsequent terms.

The recursive formula is particularly useful for teaching how sequences evolve term by term. It emphasizes how each term depends on its predecessor, illustrating the sequence's dynamic nature.