Problem 39 - Algebra and Trigonometry

Question

The common ratio in a geometric sequence is \(\frac{2}{5},\) and the fourth term is \(\frac{5}{2} .\) Find the third term.

Step-by-Step Solution

Answer

The third term is \( \frac{25}{4} \).

Step 1: Identifying the Problem

We are given a geometric sequence with a common ratio \( r = \frac{2}{5} \) and the fourth term \( a_4 = \frac{5}{2} \). We need to find the third term \( a_3 \).

Step 2: Understanding the Geometric Sequence

In a geometric sequence, each term is obtained by multiplying the previous term by the common ratio \( r \). Thus, the \( n \)-th term can be expressed as \( a_n = a_1 \cdot r^{n-1} \).

Step 3: Setting Up the Equation for the Fourth Term

Using the formula for the \( n \)-th term, we set up the equation for the fourth term: \[ a_4 = a_1 \cdot r^3 = \frac{5}{2} \].

Step 4: Expressing the Fourth Term in Terms of the Third Term

The third term can be expressed in terms of the first term: \( a_3 = a_1 \cdot r^2 \). We can also express \( a_4 \, as \, a_3 \cdot r = \frac{5}{2} \) because each term is the previous term times the common ratio.

Step 5: Solving for the Third Term

We solve for \( a_3 \) using the relationship obtained: \( a_3 \cdot r = a_4 \) or \( a_3 \cdot \frac{2}{5} = \frac{5}{2} \). Solving for \( a_3 \), we have: \[ a_3 = \frac{5}{2} \div \frac{2}{5} = \frac{5}{2} \times \frac{5}{2} = \frac{25}{4} \].

Step 6: Conclusion

Thus, the third term of the geometric sequence is \( \frac{25}{4} \).

Key Concepts

Understanding the Common Ratio

In a geometric sequence, the **common ratio** is a key concept that determines how the sequence progresses from one term to the next. This ratio, usually denoted by \( r \), indicates how each term is derived by multiplying the previous term by \( r \). In simple terms, you can imagine the common ratio as a multiplier that transforms one number into the next in a seamless manner.

For example, consider a sequence where the common ratio is \( \frac{3}{2} \). From an initial term of 2, applying the common ratio would result in subsequent terms like 3 (since \( 2 \times \frac{3}{2} = 3 \)), then 4.5 and so on. In the given exercise, the common ratio is \( \frac{2}{5} \), showing that each term is smaller than the previous one. This notion of a consistent multiplier makes the common ratio a foundational component of geometric sequences.

Exploring the Nth Term Formula

The **nth term formula** in geometric sequences allows you to calculate any term in the sequence without the need to derive all previous terms. This formula is expressed as \( a_n = a_1 \cdot r^{n-1} \), where \( a_n \) is the term you want to find, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

This compact formula is incredibly efficient for sequence problems. For example, if you need the 10th term in a sequence with \( a_1 = 2 \) and \( r = 3 \), plug these into the formula to get \( a_{10} = 2 \cdot 3^9 \). This eliminates the often cumbersome process of multiplying the ratio repeatedly. In the provided exercise, we utilize the nth term formula to figure out unknown terms given certain conditions, like the common ratio and an existing term.

Deciphering the Geometric Series

A **geometric series** is the sum of the terms of a geometric sequence. When you sum up the terms in a sequence, whether finite or infinite, you obtain a geometric series. This concept is pivotal, particularly in mathematical finance, physics, and other applications, where series sum to a particular total.

The formula for the sum of a finite geometric series consisting of \( n \) terms is given by \( S_n = a_1 \frac{1-r^n}{1-r} \), provided \( r eq 1 \). Understanding this formula helps in calculating the sum efficiently. For infinite series, where \| r \| is less than 1, the sum is \( S = \frac{a_1}{1-r} \).

In our exercise context, while the problem focuses on finding a specific term, remember that knowing about geometric series provides a broader understanding of how sequences collectively form meaningful wholes through their sums.