Problem 21 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

\(a_1 = 125, r = \frac{1}{5}\).

Step 1: Understand the given terms

We're given that \(a_3 = 5\) and \(a_8 = \frac{1}{625}\). In a geometric sequence, any term \(a_n\) is defined by \(a_n = a_1 \, r^{n-1}\). Here, \(a_3\) and \(a_8\) are the third and eighth terms, respectively.

Step 2: Express the terms in terms of \(a_1\) and \(r\)

For the third term, we have:\[ a_3 = a_1 \, r^2 = 5. \]For the eighth term, we have:\[ a_8 = a_1 \, r^7 = \frac{1}{625}. \]

Step 3: Solve for \(a_1\) using a system of equations

Express \(a_1\) from the equation for \(a_3\):\[ a_1 = \frac{5}{r^2}. \]Substitute \(a_1\) in the equation for \(a_8\):\[ \frac{5}{r^2} \, r^7 = \frac{1}{625}. \]This simplifies to:\[ 5 \, r^5 = \frac{1}{625}. \]

Step 4: Solve for \(r\)

Divide both sides by 5:\[ r^5 = \frac{1}{3125}. \]Recognize that \(3125 = 5^5\), so \[ r^5 = 5^{-5}. \]Thus, solving for \(r\) gives:\[ r = \frac{1}{5}. \]

Step 5: Find \(a_1\)

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

Step 6: Verify the solution

Check if the values satisfy the original terms. For \(a_3\):\[ a_3 = 125 \times \left(\frac{1}{5}\right)^2 = 125 \times \frac{1}{25} = 5. \]For \(a_8\):\[ a_8 = 125 \times \left(\frac{1}{5}\right)^7 = 125 \times \frac{1}{78125} = \frac{1}{625}. \]Both are correct.

Key Concepts

Geometric Progression

A geometric progression, or geometric sequence, is an ordered list of numbers where each term after the first is obtained by multiplying the previous term by a constant factor. This constant is known as the common ratio. Geometric sequences are prevalent in mathematics due to their regular structure and the patterns they form.
For example, consider a sequence starting with term \(a_1\) and a common ratio \(r\). The terms will look like this:

• First term: \(a_1\)
• Second term: \(a_1 \times r\)
• Third term: \(a_1 \times r^2\)
• Fourth term: \(a_1 \times r^3\)

And so on. This pattern continues indefinitely, making geometric progressions both predictable and powerful tools for solving mathematical problems that involve exponential growth or decay.

Common Ratio

The common ratio in a geometric sequence is a crucial component, as it determines the behavior and scale of the sequence. It is the factor by which we multiply the previous term to get the next term. Understanding the common ratio is essential for any operations involving the sequence.
In our specific problem, the common ratio \(r\) was found to be \(\frac{1}{5}\). This means each term is \(\frac{1}{5}\) of the one before it, depicting a sequence that decreases geometrically. When solving such problems, identifying the common ratio is often one of the first steps.
Mathematically, if we know two terms, say \(a_n\) and \(a_{m}\), we can find the common ratio using the formula:
\[ r = \left( \frac{a_{m}}{a_n} \right)^{\frac{1}{m-n}} \].With this ratio, other terms in the sequence can also be calculated efficiently.

Sequence Terms

Sequence terms denote the individual elements in a progression. Each term in a geometric sequence is related to the other terms through the sequence's first term \(a_1\) and the common ratio \(r\). Understanding how sequence terms are generated allows us to manipulate and solve for unknown terms.
For our problem, we begin with given terms: \(a_3 = 5\) and \(a_8 = \frac{1}{625}\). Using the formula for any term in a geometric sequence, \(a_n = a_1 \times r^{n-1}\), we relate these terms back to \(a_1\) and \(r\).
These relationships form the backbone of our equations that help us solve for the unknown terms of the sequence. Knowing any two sequence terms and their positions can allow us to find others, using basic algebra and properties of exponents, making the sequence a powerful concept in dissecting patterns and growth behaviors.

Solving Equations

Solving equations within the context of geometric sequences involves algebraically manipulating expressions to find unknown variables like the first term or common ratio. It requires setting up equations using known terms and their properties.
In this exercise, we derived two equations from \(a_3 = 5\) and \(a_8 = \frac{1}{625}\), namely:
\[ a_3 = a_1 \times r^2\] \[ a_8 = a_1 \times r^7\]Then, by expressing \(a_1\) from one of these equations and substituting it into the other, we simplify the problem to solving for the common ratio \(r\).
Once \(r\) is found, it can be substituted back into the expression for \(a_1\), thus solving for it. This methodical approach can apply to a range of problems, illustrating the utility of algebraic manipulation to solve for unknown variables.