Problem 14 - A Graphical Approach to Precalculus with Limits

Question

CHECKING ANALYTIC SKILLS Find \(a_{5}\) and \(a_{n}\) for each geometric sequence. Do not use a calculator. $$\frac{1}{2}, \frac{2}{3}, \frac{8}{9}, \frac{32}{27}, \ldots$$

Step-by-Step Solution

Answer

\(a_5 = \frac{128}{81}\), \(a_n = \frac{1}{2} \times \left(\frac{4}{3}\right)^{n-1}\).

Step 1: Understand the Geometric Sequence

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (denoted as \(r\)). We are given the sequence \(\frac{1}{2}, \frac{2}{3}, \frac{8}{9}, \frac{32}{27}, \ldots\).

Step 2: Identify the First Term and Common Ratio

The first term of the sequence, denoted as \(a_1\), is \(\frac{1}{2}\). To find the common ratio \(r\), divide the second term by the first term: \(r = \frac{2/3}{1/2} = \frac{2}{3} \times \frac{2}{1} = \frac{4}{3}\).

Step 3: Derive the General Formula for the Sequence

The general term for a geometric sequence can be expressed as \(a_n = a_1 \times r^{n-1}\). With \(a_1 = \frac{1}{2}\) and \(r = \frac{4}{3}\), the formula becomes \(a_n = \frac{1}{2} \times \left(\frac{4}{3}\right)^{n-1}\).

Step 4: Find the Fifth Term \(a_5\)

Substitute \(n = 5\) into the general formula: \(a_5 = \frac{1}{2} \times \left(\frac{4}{3}\right)^{4}\). Calculate step-by-step: \(\left(\frac{4}{3}\right)^{4} = \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} = \frac{256}{81}\). Then, \(a_5 = \frac{1}{2} \times \frac{256}{81} = \frac{256}{162}\), which simplifies to \(\frac{128}{81}\).

Step 5: State the General Term \(a_n\)

In Step 3, we found that the formula for the general term is \(a_n = \frac{1}{2} \times \left(\frac{4}{3}\right)^{n-1}\). This will generate any term in the sequence.

Key Concepts

Understanding the Common Ratio

The concept of a common ratio is key to understanding geometric sequences. A common ratio, often denoted as \( r \), is a fixed number that each term in the sequence is multiplied by to get to the next term. This multiplication happens consistently throughout the sequence, making it a uniform progression. To determine the common ratio, simply divide any term in the sequence by the previous term.
In the example provided, the sequence begins with \( \frac{1}{2} \), followed by \( \frac{2}{3} \), \( \frac{8}{9} \), and so on. To find the common ratio:

• Take the second term: \( \frac{2}{3} \).
• Divide it by the first term: \( \frac{1}{2} \).
• The calculation \( \frac{2/3}{1/2} \) simplifies to \( \frac{4}{3} \).

This ratio of \( \frac{4}{3} \) remains constant across the sequence, ensuring the same multiplication factor is applied to subsequent terms.

Exploring the General Term of a Sequence

Finding the general term in a geometric sequence allows us to calculate any term's value without having to list all prior terms. This general term, \( a_n \), is formulated using the first term and the common ratio.
The general term of a geometric sequence can be described using the formula:\[ a_n = a_1 \times r^{n-1} \]Here:

• \( a_1 \) is the first term.
• \( r \) is the common ratio.
• \( n \) represents the position of the term in the sequence.

In the exercise, with \( a_1 = \frac{1}{2} \) and \( r = \frac{4}{3} \), the general term becomes:\[a_n = \frac{1}{2} \times \left(\frac{4}{3}\right)^{n-1}\]This allows any term within the sequence to be generated by substituting the desired term's position for \( n \). This flexibility is integral to solving complex sequence problems with ease.

Decoding the Geometric Progression Formula

The geometric progression formula provides a systematic way to understand the behavior of any geometric sequence. This formula expresses how each term is derived from the previous term using multiplication by a constant, the common ratio.
The formula, \( a_n = a_1 \times r^{n-1} \), gives us insight into two critical aspects:

• The starting point of the sequence, \( a_1 \), which is the initial term.
• The exponential nature of the progression, as denoted by \( r^{n-1} \), representing repeated multiplication by the common ratio.

In a geometric sequence, every step forward involves multiplying by \( r \), showcasing the power of exponential growth or decay depending on whether \( r \) is greater or less than one.
This formula is vital in various mathematical fields, from algebra to calculus, and powers countless applications, including financial calculations such as compound interest. The elegance of the geometric progression lies in its simplicity and broad utility, making it a fundamental concept in sequential mathematics.