Problem 38
Question
Find the next two terms of each geometric sequence. $$ \frac{5}{2}, \frac{5}{3}, \frac{10}{9}, \dots $$
Step-by-Step Solution
Verified Answer
The next two terms are \( \frac{20}{27} \) and \( \frac{40}{81} \).
1Step 1: Identify the Common Ratio
A geometric sequence is defined by a constant ratio between consecutive terms. To find this ratio, divide the second term by the first term: \( r = \frac{5/3}{5/2} = \frac{5}{3} \times \frac{2}{5} = \frac{2}{3} \).
2Step 2: Verify the Common Ratio
Check if the same ratio applies between the third and second terms: \( \frac{10/9}{5/3} = \frac{10}{9} \times \frac{3}{5} = \frac{2}{3} \). Since it's consistent, the common ratio is indeed \( \frac{2}{3} \).
3Step 3: Calculate the Fourth Term
Multiply the third term by the common ratio to find the fourth term: \( 10/9 \times 2/3 = \frac{20}{27} \).
4Step 4: Calculate the Fifth Term
Multiply the fourth term by the common ratio to find the fifth term: \( 20/27 \times 2/3 = \frac{40}{81} \).
Key Concepts
Common Ratio in Geometric SequencesUnderstanding Consecutive TermsTerm Calculation in Geometric SequencesSequence Continuation by Extending the Pattern
Common Ratio in Geometric Sequences
The common ratio is a key feature of geometric sequences and is essential for understanding their progression. In such sequences, the ratio between any two consecutive terms is always the same. This constant ratio is what we call the "common ratio." Determining the common ratio is simple. You take any term in the sequence and divide it by the term preceding it. In our example sequence \( \frac{5}{2}, \frac{5}{3}, \frac{10}{9}, \dots \), we calculated the common ratio by dividing the second term \( \frac{5}{3} \) by the first term \( \frac{5}{2} \), giving us a common ratio of \( \frac{2}{3} \). This is a crucial initial step as it allows us to understand how each term relates to its predecessor and how the sequence is built.
Understanding Consecutive Terms
Consecutive terms in a geometric sequence are simply the terms that follow one after another in the order they appear. Each term is related to the one before it through the multiplication of the common ratio. For example, in the sequence \( \frac{5}{2}, \frac{5}{3}, \frac{10}{9}, \dots \), each term is derived by multiplying the previous term by the common ratio of \( \frac{2}{3} \).
- The second term \( \frac{5}{3} \) is obtained by multiplying the first term \( \frac{5}{2} \) by \( \frac{2}{3} \).
- The third term \( \frac{10}{9} \) results from multiplying the second term by \( \frac{2}{3} \).
Term Calculation in Geometric Sequences
Term calculation is all about finding the unknown terms once you have the common ratio. This is where understanding prior calculations comes in very handy. To calculate any term in the sequence, you simply multiply the preceding term by the common ratio. For the given sequence:
- The fourth term can be calculated by multiplying the third term \( \frac{10}{9} \) by the common ratio \( \frac{2}{3} \) to get \( \frac{20}{27} \).
- Similarly, the fifth term results from multiplying the fourth term \( \frac{20}{27} \) also by \( \frac{2}{3} \), which gives \( \frac{40}{81} \).
Sequence Continuation by Extending the Pattern
Sequence continuation involves extending the pattern of a geometric sequence to find future terms. With the common ratio in hand, this task becomes straightforward. After calculating the fourth and fifth terms for our example sequence, continuing further would follow the same method:
- Take the known term, in this case, the fifth term \( \frac{40}{81} \), and multiply it by the common ratio \( \frac{2}{3} \) to find the sixth term.
- Repeating this step allows the sequence to grow indefinitely, offering a precise view of its uninterrupted pattern.
Other exercises in this chapter
Problem 38
Find the sum of each infinite geometric series, if it exists. \(1-0.5+0.25-\ldots\)
View solution Problem 38
Find \(S_{n}\) for each geometric series described. $$ a_{1}=162, r=\frac{1}{3}, n=6 $$
View solution Problem 38
Write an equation for the nth term of each arithmetic sequence. \(18,11,4,-3, \ldots\)
View solution Problem 38
Find the sum of each arithmetic series. $$ \sum_{n=1}^{6}(2 n+11) $$
View solution