Problem 6 - Precalculus Mathematics for Calculus

Question

The \(n\) th term of a sequence is given. (a) Find the first five terms of the sequence. (b) What is the common ratio \(r ?\) (c) Graph the terms you found in (a). $$a_{n}=3(-4)^{n-1}$$

Step-by-Step Solution

Answer

First five terms: 3, -12, 48, -192, 768. Common ratio is -4.

Step 1: Find the first term

To find the first term of the sequence, substitute \(n = 1\) into the given formula: \[a_1 = 3(-4)^{1-1} = 3(-4)^{0} = 3 \cdot 1 = 3\]. Thus, the first term is 3.

Step 2: Find the second term

Substitute \(n = 2\) into the formula to find the second term: \[a_2 = 3(-4)^{2-1} = 3(-4)^{1} = 3 \cdot (-4) = -12\]. The second term is -12.

Step 3: Find the third term

Now, substitute \(n = 3\) into the formula: \[a_3 = 3(-4)^{3-1} = 3(-4)^{2} = 3 \cdot 16 = 48\]. The third term is 48.

Step 4: Find the fourth term

To find the fourth term, substitute \(n = 4\) into the sequence formula: \[a_4 = 3(-4)^{4-1} = 3(-4)^{3} = 3 \cdot (-64) = -192\]. The fourth term is -192.

Step 5: Find the fifth term

Substitute \(n = 5\) into the formula to determine the fifth term: \[a_5 = 3(-4)^{5-1} = 3(-4)^{4} = 3 \cdot 256 = 768\]. The fifth term is 768.

Step 6: Calculate the common ratio

The common ratio \(r\) in a geometric sequence is found by dividing any term by the previous term. Using the second and first terms: \[r = \frac{a_2}{a_1} = \frac{-12}{3} = -4\]. Thus, the common ratio is \(-4\).

Step 7: Graph the terms

Plot the first five terms of the sequence on a graph where the x-axis represents the term number \(n\) and the y-axis represents the value of \(a_n\). The points to plot are (1, 3), (2, -12), (3, 48), (4, -192), (5, 768). The graph should show an alternating pattern due to the negative ratio.

Key Concepts

Common Ratio

In a geometric sequence, the common ratio \( r \) is the constant factor between consecutive terms. Understanding this ratio is crucial because it defines the pattern of the sequence. In the given sequence \( a_{n} = 3(-4)^{n-1} \), the common ratio is \(-4\). This means that each term is steadily growing by multiplying the previous term by \(-4\).

To find the common ratio, we take any term and divide it by the term right before it. In our solution, we used the second term \(-12\) and the first term 3 to verify this:

• Common Ratio Formula: \( r = \frac{a_2}{a_1} \)
• Calculation: \( r = \frac{-12}{3} = -4 \)

Understanding the common ratio helps predict future terms and the sequence's nature, which in this case fluctuates between positive and negative due to the negative common ratio.

Graphing Sequences

Graphing sequences provides a visual way to see how they behave. For our sequence \( a_n = 3(-4)^{n-1} \), plotting the values of the first five terms offers insights into its pattern. Each 'term number' \( n \) is plotted on the x-axis, and the 'term value' \( a_n \) on the y-axis.

Our points from (1, 3) to (5, 768) visually represent how rapidly the sequence changes:

• (1, 3)
• (2, -12)
• (3, 48)
• (4, -192)
• (5, 768)

The sequence's graph shows an alternating pattern above and below the x-axis due to the negative common ratio \(-4\).

Graphing helps in understanding growth or decay in sequences and highlights the oscillating nature when the common ratio is negative.

Recursive Formulas

Recursive formulas offer a way to describe sequences by expressing each term based on the preceding term or terms. They are particularly useful when describing sequences where calculating terms directly is cumbersome. However, in geometric sequences, like \( a_n = 3(-4)^{n-1} \), each term can also be derived recursively.

For a sequence defined as \( a_n \), the recursive formula can be written as:

• \( a_1 = 3 \) (initial term definition)
• \( a_{n} = r \times a_{n-1} \) for \( n > 1 \)

In our example, with \( r = -4 \), each term is obtained by multiplying the previous term by \(-4\).

Recursive formulas are concise for describing parts of sequences. They are especially suitable in programming and algorithms where initial terms set a foundation for repetition.