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Chapter 15: Problem 32

Find the indicated term of each geometric sequence. $$1,3,9,27, \dots, a_{10}$$

Short Answer

Expert verified
The 10th term of the geometric sequence \(1, 3, 9, 27, \dots\) is 19,683.

Step by step solution

01

Identify the common ratio

The given sequence is \(1, 3, 9, 27, \dots\). In order to find the common ratio, divide any term by its previous term. For example, divide the second term (3) by the first term (1): \(r = \frac{3}{1} = 3\) The common ratio is 3.
02

Use the formula for the nth term

The formula for the nth term of a geometric sequence is: \(a_n = a_1 * r^{n-1}\) where \(a_n\) is the nth term, \(a_1\) is the first term, r is the common ratio, and n is the number of terms.
03

Substitute the known values into the equation

We know the first term \(a_1 = 1\), the common ratio \(r = 3\), and we want to find the 10th term (n = 10). Plugging these values into the formula, we get: \(a_{10} = 1 * 3^{10-1}\)
04

Calculate the 10th term

Now, simplify the equation: \(a_{10} = 1 * 3^9\) \(a_{10} = 3^9\) \(a_{10} = 19683\) The 10th term of the given geometric sequence is 19,683.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Ratio
In a geometric sequence, the common ratio is a fundamental characteristic. It defines how the sequence progresses from one term to the next. The key idea is that each term is the product of the previous term and a fixed, constant value known as the common ratio. To find the common ratio, you can take any term in the sequence and divide it by the term before it.
Let's look at the provided sequence: 1, 3, 9, 27, and so on. Here:
  • To find the common ratio, divide the second term (3) by the first term (1): \[ r = \frac{3}{1} = 3\]
  • Similarly, divide the third term (9) by the second term (3): \[ r = \frac{9}{3} = 3\]
In both cases, we get a common ratio of 3. This means each term is three times the previous term.
Nth Term Formula
The Nth term formula in a geometric sequence is crucial for finding any term given its position. The formula allows you to calculate terms without having to individually multiply through the sequence from one term to the next.
The general formula for finding the nth term (\(a_n\)) is:\[a_n = a_1 \times r^{(n-1)}\]
  • Here, \(a_1\) is the first term of the sequence.
  • \(r\) is the common ratio, as explained earlier.
  • \(n\) is the position of the term you want to find.
For the provided sequence, let鈥檚 identify its parts to find the 10th term:
  • First term \(a_1 = 1\)
  • Common ratio \(r = 3\)
  • We seek \(a_{10}\)
Substitute these values into the formula:\[a_{10} = 1 \times 3^{(10-1)} = 3^9\]Calculate \(3^9\) to get the 10th term as 19,683.
Exponential Growth
Geometric sequences are a prime example of exponential growth, where each term increases or decreases by multiplying by a constant factor, known as the common ratio. Unlike arithmetic sequences that grow by adding a constant number, geometric sequences multiply by a constant ratio.
This means the sequence grows or decays at a rate that becomes ever-larger (or smaller) over time.
  • In the given sequence, every new number is obtained by multiplying the previous number by 3.
  • This effect is much more pronounced as the sequence progresses, as seen with lower terms like \(1, 3, 9,\) up to larger ones like 19,683 for the 10th term.
In the context of exponential growth, geometric sequences can model real-world phenomena, such as population growth, investment returns, and other scenarios where growth accelerates over time.

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Most popular questions from this chapter

Find the sum of the terms of the infinite geometric sequence, if possible. $$4,-12,36,-108, \dots$$

Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$\sum_{i=1}^{7} 9(2)^{i}$$

Use the binomial theorem to expand each expression. $$(4 c-3 d)^{4}$$

Use the binomial theorem to expand each expression. $$\left(\frac{1}{2} m-3 n\right)^{4}$$

Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}$$
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