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Chapter 8: Problem 19

Write a formula for the general term (the nth term) of each geometric sequence. Then use the formula for \(a_{n}\)to find \(a_{7},\) the seventh term of the sequence. $$18,6,2, \frac{2}{3}, \dots$$

Short Answer

Expert verified
The 7th term of the sequence is \( \frac {2}{243}\).

Step by step solution

01

Find the Common Ratio

First order of business is to find the common ratio of the geometric sequence. This can be done by dividing any term in the sequence by its previous term. For example, dividing the second term by the first term, \(r = \frac {6}{18} = \frac {1}{3}\)
02

Formulate the General Term Formula

Now that we have the common ratio, the next step is to derive the expression for the nth term of the geometric sequence. A geometric sequence can be represented generally as \(a_n = a_1 * r^{(n-1)}\), where \(a_n\) is the nth term, \(a_1\) is the first term, r is the common ratio, and n is the term number. Substituting \(a_1 = 18\) and \(r = \frac {1}{3}\), we have \(a_n = 18 * (\frac{1}{3})^{n-1}\)
03

Calculate Term a7

The final step is to calculate the 7th term \(a_7\) in the sequence. Substitute \(n = 7\) into the general term formula. Thus, \(a_7 = 18 * (\frac{1}{3})^{7-1} = \frac {2}{243}\)

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Key Concepts

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

Understanding the Common Ratio
In a geometric sequence, each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the "common ratio." This ratio is a crucial part of understanding how the sequence progresses.
If you know any two consecutive terms in a sequence, you can calculate the common ratio by dividing the second term by the first. For instance, in the sequence given in the exercise, we have the terms 18 and 6. Dividing 6 by 18 results in \( \frac{1}{3} \), indicating that each term is \( \frac{1}{3} \) of the term before it.
Knowing the common ratio helps you predict the sequence's future behavior, as each term becomes more predictable as you understand this pattern.
Exploring the nth Term Formula
The nth term formula is a key tool in identifying any term in a geometric sequence without needing to list all the terms. This formula is handy, especially when working with large sequences.
For a geometric sequence, the nth term \( a_n \) can be defined using the formula \( a_n = a_1 \cdot r^{(n-1)} \). Here, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term's position.
For the sequence in question, substituting \( a_1 = 18 \) and \( r = \frac{1}{3} \) gives us the formula \( a_n = 18 \cdot \left( \frac{1}{3} \right)^{n-1} \). This formula allows you to compute any term in the sequence with ease and precision.
How to Use the General Term of a Sequence
Using the general term formula, you can calculate any term in a geometric sequence. This is particularly useful when asked to find a specific term, such as the seventh term, without listing every term up to that point.
For example, using the formula derived \( a_n = 18 \cdot \left( \frac{1}{3} \right)^{n-1} \), to find the seventh term \( a_7 \), simply substitute 7 for \( n \).
This leads to \( a_7 = 18 \cdot \left( \frac{1}{3} \right)^{6} \). Calculating this step-by-step ensures accuracy:
  • Calculate \( \left( \frac{1}{3} \right)^{6} = \frac{1}{729} \)
  • Multiply \( 18 \cdot \frac{1}{729} = \frac{18}{729} = \frac{2}{243} \)
Thus, the seventh term \( a_7 \) is \( \frac{2}{243} \), showcasing how the general term formula simplifies finding specific terms.

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Most popular questions from this chapter

Follow the outline on the next page to use mathematical induction to prove that $$ \begin{aligned}(a+b)^{n}=\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\\+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned} $$ a. Verify the formula for \(n=1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$ \begin{aligned}&(a+b)^{k+1}=\left(\begin{array}{l}k \\\0\end{array}\right) a^{k+1}+\left[\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)\right] a^{k} b\\\&\begin{array}{l}+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1} \end{array}\end{aligned} $$ e. Use the result of Exercise 74 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\\ r+1\end{array}\right)$$=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\) then \(\left(\begin{array}{l}k \\ 0\end{array}\right)+\left(\begin{array}{l}k \\\ 1\end{array}\right)=\left(\begin{array}{c}k+1 \\\1\end{array}\right)\) and\(\left(\begin{array}{l}k \\ 1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)=\left(\begin{array}{c}k+1 \\ 2\end{array}\right)\) f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)=\left(\begin{array}{c}k+1 \\ 0\end{array}\right) \quad\) (why?) and \(\left(\begin{array}{l}k \\ k\end{array}\right)=\) \(\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$ (x-2 y)^{10} $$

Show that $$ \left(\begin{array}{l}n \\\r\end{array}\right)+\left(\begin{array}{c}n \\\r+1\end{array}\right)=\left(\begin{array}{l}n+1 \\\r+1 \end{array}\right) $$ Hints: $$ \begin{aligned}&(n-r) !=(n-r)(n-r-1) !\\\&(r+1) !=(r+1) r !\end{aligned} $$

The president of a large company with \(10,000\) employees is considering mandatory cocaine testing for every employee. The test that would be used is \(90 \%\) accurate, meaning that it will detect \(90 \%\) of the cocaine users who are tested, and that \(90 \%\) of the nonusers will test negative. This also means that the test gives \(10 \%\) false positive. Suppose that \(1 \%\) of the employees actually use cocaine. Find the probability that someone who tests positive for cocaine use is, indeed, a user. What does this probability indicate in terms of the percentage of employees who test positive who are not actually users? Discuss these numbers in terms of the issue of mandatory drug testing. Write a paper either in favor of or against mandatory drug testing, incorporating the actual percentage accuracy for such tests.

Evaluate the given binomial coefficient. $$ \left(\begin{array}{l}6 \\\6\end{array}\right) $$
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