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Chapter 9: Problem 26

Writing the \(n\) th Term of a Geometric Sequence, write the first five terms of the geometric sequence. Determine the common ratio and write the \(n\) th term of the sequence as a function of \(n .\) $$ a_{1}=5, \quad a_{k+1}=-2 a_{k} $$

Short Answer

Expert verified
The first five terms of the geometric sequence are \(5, -10, 20, -40, 80\). The common ratio is \(-2\) and the \(n\)th term of the sequence as function of \(n\) is \(a_n = 5*(-2)^{n-1}\).

Step by step solution

01

Determine the First Five Terms

Use the given common ratio and the first term to determine the subsequent terms. This involves multiplying each previous term by \(-2\). So, the first five terms are: \(a_1 = 5, a_2 = -2*5 = -10, a_3 = -2*(-10) = 20, a_4 = -2*20 = -40, a_5 = -2*(-40) = 80.\)
02

Determine the Common Ratio

As given in the sequence rule, the common ratio, \(r\), is \(-2\). In a geometric sequence, the ratio is the constant value we multiply to get from one term to the next.
03

Write the \(n\)th Term of the Sequence

The \(n\)th term in a geometric sequence is determined as \(a_n = a_1*r^{(n-1)}\). Here, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. Substituting the given values, we get \(a_n = 5*(-2)^{n-1}\). Thus, the \(n\)th term of the sequence as function of \(n\) is \(a_n = 5*(-2)^{n-1}\).

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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, understanding the common ratio is key to identifying how the sequence progresses. The common ratio, often represented as \( r \), is the factor by which we multiply a term in order to get the next term. It's a constant value that helps in maintaining the consistent pattern in the sequence.
  • To find the common ratio, you can divide any term by the previous term.
  • In this exercise, the common ratio \( r \) given is \(-2\).
  • This means each term is obtained by multiplying the previous term by \(-2\).
For instance, starting with the term \( a_1 = 5 \), we multiply by \(-2\) to get \( a_2 = -10 \), and so on. Identifying this ratio is crucial, as it lays the groundwork for calculating any term in the sequence.
Nth Term Formula
The nth term formula in a geometric sequence allows you to find any term in the sequence without listing all of the preceding terms. It is expressed using the first term and common ratio.The formula for the nth term \( a_n \) of a geometric sequence is given by:\[ a_n = a_1 \times r^{(n-1)} \]
  • \( a_1 \) is the first term.
  • \( r \) is the common ratio.
  • \( n \) is the position of the term in the sequence.
For example, in our sequence, the given values are \( a_1 = 5 \) and \( r = -2 \). Substituting these into the formula gives:\[ a_n = 5 \times (-2)^{(n-1)} \]This function tells us that to find any specific term, like the 3rd or 10th, we simply plug the term number into this formula.
First Five Terms
Calculating the first five terms of a geometric sequence helps in understanding the sequence鈥檚 initial behavior and verifying the correctness of the common ratio and formula.Starting with the first term, \( a_1 = 5 \), the first five terms are calculated by continuously multiplying by the common ratio \(-2\):
  • The first term is \( 5 \).
  • Multiply by \(-2\) to get the second term, \(-10\).
  • The third term is \( 20 \), after multiplying \(-10\) by \(-2\).
  • Repeat the multiplication to get the fourth term, \(-40\).
  • Finally, multiply by \(-2\) to obtain the fifth term, \( 80 \).
These calculations help reinforce the application of the sequence鈥檚 rules, providing a practical understanding of how each term is derived based on its predecessors.

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Most popular questions from this chapter

Finding a Sum In Exercises \(45-54\) , find the sum using the formulas for the sums of powers of integers. $$\sum_{n=1}^{6} n^{2}$$

Finding a Quadratic Model In Exercises \(69-74\) find a quadratic model for the sequence with the indicated terms. $$a_{0}=3, a_{1}=3, a_{4}=15$$

Finding a Sum In Exercises \(45-54\) , find the sum using the formulas for the sums of powers of integers. $$\sum_{j=1}^{10}\left(3-\frac{1}{2} j+\frac{1}{2} j^{2}\right)$$

Finding a Formula for a Sum In Exercises \(41-44\) , use mathematical induction to find a formula for the sum of the first \(n\) terms of the sequence. $$1,5,9,13, \dots$$

Consider a group of \(n\) people. (a) Explain why the following pattern gives the probabilities that the \(n\) people have distinct birthdays. $$\begin{array}{l}{n=2 : \frac{365}{365} \cdot \frac{364}{365}=\frac{365 \cdot 364}{365^{2}}} \\ {n=3 : \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}=\frac{365 \cdot 364 \cdot 363}{365^{3}}}\end{array}$$ (b) Use the pattern in part (a) to write an expression for the probability that \(n=4\) people have distinct birthdays. (c) Let \(P_{n}\) be the probability that the \(n\) people have distinct birthdays. Verify that this probability can be obtained recursively by $$P_{1}=1\( and \)P_{n}=\frac{365-(n-1)}{365} P_{n-1}$$ (d) Explain why \(Q_{n}=1-P_{n}\) gives the probability that at least two people in a group of \(n\) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \(\frac{1}{2} ?\) Explain.
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