Problem 19 - Beginning and Intermediate Algebra

Question

Find the general term, \(a_{m}\) for each geometric sequence. Then, find the indicated term. $$a_{1}=2, r=\frac{1}{5} ; a_{4}$$

Step-by-Step Solution

Answer

The general term for this geometric sequence is given by \(a_m = 2 \cdot \left(\frac{1}{5}\right)^{(m-1)}\), and the 4th term, \(a_4\) is \(\frac{2}{125}\).

Step 1: Recall and apply the formula for the nth term of a geometric sequence

Given the formula \(a_m = a_1 \cdot r^{(m-1)}\), we'll use the given values for \(a_1\) and \(r\) to find the general term of the sequence. For this problem, we know: $$a_1 = 2, \quad r = \frac{1}{5}$$ Substituting these values into the formula, we get: $$a_m = 2 \cdot \left(\frac{1}{5}\right)^{(m-1)}$$

Step 2: Find the 4th term, \(a_4\)

Now that we have the general term, \(a_m\), we can find \(a_4\) by substituting \(m = 4\) into the formula: $$a_4 = 2 \cdot \left(\frac{1}{5}\right)^{(4-1)}$$ Simplify the expression in the exponent: $$a_4 = 2 \cdot \left(\frac{1}{5}\right)^{3}$$

Step 3: Calculate \(a_4\)

Now we'll calculate \(a_4\) using the simplified formula: $$a_4 = 2 \cdot \frac{1}{5^3} = 2 \cdot \frac{1}{125}$$ Finally, we'll simplify the fraction to get \(a_4\): $$a_4 = \frac{2}{125}$$ The general term for this geometric sequence is given by \(a_m = 2 \cdot \left(\frac{1}{5}\right)^{(m-1)}\), and the 4th term, \(a_4\) is \(\frac{2}{125}\).

Key Concepts

Understanding the nth term formula

The nth term formula is a fundamental building block in the study of geometric sequences. A geometric sequence is a list 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. To find any term in this sequence, we use the nth term formula:

• \( a_m = a_1 \cdot r^{(m-1)} \)

This formula allows you to calculate the value of the \(m\)-th term, \(a_m\), without having to list all previous terms. Here, \(a_1\) is the first term of the sequence, \(r\) is the common ratio, and \(m\) is the term position we want to find.
By knowing the first term and the common ratio, you can plug these into the formula to compute any term in the sequence directly.

Exploring the general term of a sequence

The general term of a sequence is a formula that represents the \(m\)-th term as a function of \(m\). For the given geometric sequence where \(a_1 = 2\) and \(r = \frac{1}{5}\), the general term is found using the nth term formula:

• \( a_m = 2 \cdot \left(\frac{1}{5}\right)^{(m-1)} \)

This expression gives us a way to calculate any term in the sequence, not just the initial few terms we might easily observe.
The general term formula includes the independent variable \(m\), meaning it equates a specific term in the sequence with its position index, \(m\). Understanding this formula allows us to explore sequences to any level without listing each of the terms manually, making it a powerful tool in mathematics.

Decoding the common ratio

The common ratio is a key component in a geometric sequence. It determines how each term relates to its preceding term. In this sequence, the common ratio \(r\) is given as \(\frac{1}{5}\). This means that to get from one term to the next in this sequence, you multiply the current term by \(\frac{1}{5}\).
The common ratio must be constant throughout a geometric sequence. It can be a fraction, whole number, or even negative, but it cannot be zero as this would make the sequence degenerate. By understanding the common ratio, you can predict how the sequence is expanding or contracting. If \(|r| < 1\), as in our example, the terms will decrease in absolute value, approaching zero as we calculate higher terms.

Explaining exponentiation in sequences

Exponentiation in sequences refers to raising the common ratio \(r\) to a power, as seen in the nth term formula. This operation is pivotal in finding terms in a geometric sequence.
For the sequence \(a_m = 2 \cdot \left(\frac{1}{5}\right)^{(m-1)}\), the \((m-1)\) exponent denotes how many times the common ratio \(r\) is multiplied by itself. This captures the cumulative effect of applying the ratio successively to reach the term of interest.

• For example, for \(a_4\), we use \((4-1)\) as the exponent, meaning \(\frac{1}{5}\) is multiplied by itself three times resulting in \( \left(\frac{1}{5}\right)^3 = \frac{1}{125} \).

Your understanding of exponentiation consolidates your ability to apply the nth term formula and find any specific term in a geometric sequence with confidence.