Problem 24 - Algebra for College Students

Question

Find the common ratio of the geometric sequence with 2nd term \(\frac{8}{3}\) and 5 th term \(\frac{64}{81} . \quad \frac{2}{3}\)

Step-by-Step Solution

Answer

The common ratio is \( \frac{2}{3} \).

Step 1: Understand the problem

We are given the 2nd term and the 5th term of a geometric sequence and need to find the common ratio. In a geometric sequence, each term is found by multiplying the previous term by the common ratio.

Step 2: Write the formula for the nth term

The nth term of a geometric sequence is given by \( a_n = a_1 imes r^{n-1} \) where \( a_1 \) is the first term and \( r \) is the common ratio.

Step 3: Set up equations for given terms

For the 2nd term \( \left( a_2 = \frac{8}{3} \right) \), we have \( a_1 \times r = \frac{8}{3} \).For the 5th term \( \left( a_5 = \frac{64}{81} \right) \), we have \( a_1 \times r^4 = \frac{64}{81} \).

Step 4: Solve for \( r \)

Divide the equation for \( a_5 \) by the equation for \( a_2 \):\[ \frac{a_1 r^4}{a_1 r} = \frac{64}{81} \div \frac{8}{3} \]Simplify the left side:\[ r^3 = \frac{64}{81} \times \frac{3}{8} \]\[ r^3 = \frac{64 \times 3}{81 \times 8} \]\[ r^3 = \frac{192}{648} \]Simplify to get:\[ r^3 = \frac{8}{27} \]Taking the cube root, we find:\[ r = \frac{2}{3} \]

Step 5: Verify the solution

Verify by calculating using the common ratio. If the 2nd term is \( \frac{8}{3} \), the 5th term should satisfy:\[ a_2 \times r^3 = \frac{8}{3} \times \left( \frac{2}{3} \right)^3 \]Calculate \( \left( \frac{2}{3} \right)^3 = \frac{8}{27} \), then:\[ \frac{8}{3} \times \frac{8}{27} = \frac{64}{81} \]As this equals the 5th term given, our solution is correct.

Key Concepts

Common Ratio

In a geometric sequence, one of the key elements is the common ratio, denoted as \( r \). It is the factor by which you multiply each term to get the next term in the sequence. Finding this consistent multiplier is crucial for understanding how the sequence progresses. In our example, we needed to find \( r \) given specific terms of the sequence.
Let's break it down:

• We know from the exercise that the 2nd term is \( \frac{8}{3} \).
• The 5th term is given as \( \frac{64}{81} \).

The formula used to find any term in a geometric sequence involves the common ratio, and without it, you can't predict future or past terms effectively. We used the provided terms to solve for \( r \), confirming that each term is indeed just the previous term multiplied by \( \frac{2}{3} \). This consistent multiplication maintains the structure and pattern of the geometric sequence throughout.

Nth Term Formula

The nth term formula is a powerful tool in geometric sequences, offering a precise way to find any term's value based solely on its position within the sequence. This formula is expressed as follows:
\[ a_n = a_1 \times r^{n-1} \]

• \( a_n \) represents the nth term you're trying to determine.
• \( a_1 \) is the first term of the sequence.
• \( r \) is the common ratio, which we found to be \( \frac{2}{3} \) in our problem.

The elegance of this formula is its simplicity and the clarity it provides. In our situation, since we didn't know \( a_1 \) directly, we used known terms to solve equations. This step helped us understand how each term evolves based on its predecessor. The nth term formula effectively bridges the gap between knowing random terms and predicting other terms or the sequence's behavior.

Geometric Progression

A geometric progression, or geometric sequence, is a collection of numbers organized in such a way that the ratio between consecutive numbers stays constant. This ratio is what we refer to as the common ratio. In mathematical terms, if you begin with a base term \( a_1 \), each following term is defined as:
\( a_1, a_1r, a_1r^2, a_1r^3, \ldots \)Every subsequent term is simply the previous term multiplied by the common ratio \( r \). For instance, if your common ratio is \( \frac{2}{3} \), then each term is always \( \frac{2}{3} \) of the previous term.
This consistent multiplication not only defines the geometric progression but also ensures that the sequence follows a predictable pattern. Understanding this concept is vital, as it allows you to reconstruct parts of a sequence from just a few initial pieces, as we did in the exercise by using the 2nd and 5th terms to find the common ratio.