Problem 17 - A Graphical Approach to College Algebra

Question

Find \(a_{1}\) and \(r\) for each geometric sequence. $$a_{3}=5, a_{8}=\frac{1}{625}$$

Step-by-Step Solution

Answer

The first term \(a_{1}\) is 125, and the common ratio \(r\) is \(\frac{1}{5}\).

Step 1: Understanding the Problem

We are given two terms, \(a_3 = 5\) and \(a_8 = \frac{1}{625}\), of a geometric sequence. Our goal is to find the first term \(a_1\) and the common ratio \(r\) of the sequence.

Step 2: Writing the General Formula

The formula for the \(n\)-th term of a geometric sequence is given by \(a_n = a_1 \, r^{n-1}\). We will use this formula for \(a_3\) and \(a_8\).

Step 3: Setting up Equations

Using the formula, we have two equations:1. \(a_3 = a_1 \, r^{2} = 5\)2. \(a_8 = a_1 \, r^{7} = \frac{1}{625}\).

Step 4: Expressing \(a_1\) from \(a_3\)

From the first equation \(a_1 \, r^{2} = 5\), express \(a_1\) as \(a_1 = \frac{5}{r^{2}}\).

Step 5: Substitute \(a_1\) into the Second Equation

Substitute \(a_1\) from Step 4 into the second equation:\[\frac{5}{r^{2}} \, r^{7} = \frac{1}{625}\].

Step 6: Simplify the Second Equation

Simplify the equation:\[5 \, r^{5} = \frac{1}{625}\] Divide both sides by 5:\[r^{5} = \frac{1}{3125}\].

Step 7: Solve for \(r\)

Taking the fifth root of both sides:\[r = \left(\frac{1}{3125}\right)^{\frac{1}{5}} = \frac{1}{5}\].

Step 8: Find \(a_1\) Using \(r\)

Substitute \(r = \frac{1}{5}\) back into the expression for \(a_1:\)\[a_1 = \frac{5}{\left(\frac{1}{5}\right)^{2}} = \frac{5}{\frac{1}{25}} = 5 \times 25 = 125\].

Key Concepts

Common Ratio

In a geometric sequence, the **common ratio** is the factor by which each term is multiplied to obtain the next term. It's a key element that defines the entire sequence. Understanding the common ratio is essential to solving problems involving geometric sequences.

To find the common ratio, you can use the formula :

• Look at any two consecutive terms; let’s call them \(a_n\) and \(a_{n+1}\).
• Use the equation: \(r = \frac{a_{n+1}}{a_n}\).

In the problem, we have identified this ratio based on non-consecutive terms \(a_3\) and \(a_8\). By setting up equations that incorporate these terms, we calculated \(r = \frac{1}{5}\). Computing the common ratio allows us to unravel many properties of the sequence, like easily finding any term or analyzing its behavior.

Geometric Series

A **geometric series** is the sum of the terms in a geometric sequence. It is important in mathematics because it allows us to find the sum of potentially infinitely many terms.However, we focus on finding specific terms within this series in this context.

• Unlike an arithmetic series, which grows by addition, a geometric series grows by multiplication, making the rate of increase or decrease much faster.
• The sum formula for a finite geometric series with \(n\) terms is given by: \[ S_n = a_1 \frac{1-r^n}{1-r} \quad \text{(if } r eq 1\text{)} \] where \(a_1\) is the first term, and \(r\) is the common ratio.

In addition to determining individual terms like \(a_1\) and the common ratio, understanding how these relate to the entire series allows for deeper insights, including the convergence of infinite series in more advanced studies. Even though initially we didn't find the sum of the sequence, knowing the first term and common ratio provides the basis for calculating such a value in practical problems.

Nth Term Formula

The **Nth Term Formula** of a geometric sequence is essential because it allows us to find any term in the sequence without listing all prior terms. This is especially useful in situations involving large sequences or finding terms far removed from the initial terms.

In the formula:

• \(a_n = a_1 \cdot r^{n-1}\)
• \(a_n\) represents the term at position \(n\), \(a_1\) is the initial term, and \(r\) is the common ratio.

This formula is powerful and versatile. Using \(a_3 = 5\) and \(a_8 = \frac{1}{625}\) in our problem, we set two separate equations based on this formula to ultimately find \(a_1 = 125\) and \(r = \frac{1}{5}\).

Thus, the formula not only helps find specific terms but also aids in solving unknowns within the relationship between terms, as seen with our derived values.