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9.3: Geometric Sequences and Series

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Skills to Develop

• Identify the common ratio of a geometric sequence.
• Find a formula for the general term of a geometric sequence.
• Calculate the $$n$$th partial sum of a geometric sequence.
• Calculate the sum of an infinite geometric series when it exists.

Geometric Sequences

A geometric sequence18, or geometric progression19, is a sequence of numbers where each successive number is the product of the previous number and some constant $$r$$.

$$a_{n}=r a_{n-1} \quad\color{Cerulean}{Geometric\:Sequence}$$

And because $$\frac{a_{n}}{a_{n-1}}=r$$, the constant factor $$r$$ is called the common ratio20. For example, the following is a geometric sequence,

$$9,27,81,243,729 \ldots$$

Here $$a_{1} = 9$$ and the ratio between any two successive terms is $$3$$. We can construct the general term $$a_{n}=3 a_{n-1}$$ where,

\begin{aligned} a_{1} &=9 \\ a_{2} &=3 a_{1}=3(9)=27 \\ a_{3} &=3 a_{2}=3(27)=81 \\ a_{4} &=3 a_{3}=3(81)=243 \\ a_{5} &=3 a_{4}=3(243)=729 \\ & \vdots \end{aligned}

In general, given the first term $$a_{1}$$ and the common ratio $$r$$ of a geometric sequence we can write the following:

\begin{aligned} a_{2} &=r a_{1} \\ a_{3} &=r a_{2}=r\left(a_{1} r\right)=a_{1} r^{2} \\ a_{4} &=r a_{3}=r\left(a_{1} r^{2}\right)=a_{1} r^{3} \\ a_{5} &=r a_{3}=r\left(a_{1} r^{3}\right)=a_{1} r^{4} \\ & \vdots \end{aligned}

From this we see that any geometric sequence can be written in terms of its first element, its common ratio, and the index as follows:

$$a_{n}=a_{1} r^{n-1} \quad\color{Cerulean}{Geometric\:Sequence}$$

In fact, any general term that is exponential in $$n$$ is a geometric sequence.

Example $$\PageIndex{1}$$:

Find an equation for the general term of the given geometric sequence and use it to calculate its $$10^{th}$$ term: $$3, 6, 12, 24, 48…$$

Solution

Begin by finding the common ratio,

$$r=\frac{6}{3}=2$$

Note that the ratio between any two successive terms is $$2$$. The sequence is indeed a geometric progression where $$a_{1} = 3$$ and $$r = 2$$.

\begin{aligned} a_{n} &=a_{1} r^{n-1} \\ &=3(2)^{n-1} \end{aligned}

Therefore, we can write the general term $$a_{n}=3(2)^{n-1}$$ and the $$10^{th}$$ term can be calculated as follows:

\begin{aligned} a_{10} &=3(2)^{10-1} \\ &=3(2)^{9} \\ &=1,536 \end{aligned}

$$a_{n}=3(2)^{n-1} ; a_{10}=1,536$$

The terms between given terms of a geometric sequence are called geometric means21.

Example $$\PageIndex{2}$$:

Find all terms between $$a_{1} = −5$$ and $$a_{4} = −135$$ of a geometric sequence. In other words, find all geometric means between the $$1^{st}$$ and $$4^{th}$$ terms.

Solution

Begin by finding the common ratio $$r$$. In this case, we are given the first and fourth terms:

\begin{aligned} a_{n} &=a_{1} r^{n-1} \quad\color{Cerulean} { Use \: n=4} \\ a_{4} &=a_{1} r^{4-1} \\ a_{4} &=a_{1} r^{3} \end{aligned}

Substitute $$a_{1} = −5$$ and $$a_{4} = −135$$ into the above equation and then solve for $$r$$.

\begin{aligned}-135 &=-5 r^{3} \\ 27 &=r^{3} \\ 3 &=r \end{aligned}

Next use the first term $$a_{1} = −5$$ and the common ratio $$r = 3$$ to find an equation for the $$n$$th term of the sequence.

\begin{aligned} a_{n} &=a_{1} r^{n-1} \\ a_{n} &=-5(3)^{n-1} \end{aligned}

Now we can use $$a_{n}=-5(3)^{n-1}$$ where $$n$$ is a positive integer to determine the missing terms.

$$\left.\begin{array}{l}{a_{1}=-5(3)^{1-1}=-5 \cdot 3^{0}=-5} \\ {a_{2}=-5(3)^{2-1}=-5 \cdot 3^{1}=-15} \\ {a_{3}=-5(3)^{3-1}=-5 \cdot 3^{2}=-45} \\ a_{4}=-5(3)^{4-1}=-5\cdot3^{3}=-135\end{array}\right\} \color{Cerulean}{geometric\:means}$$

$$-15,-45$$

The first term of a geometric sequence may not be given.

Example $$\PageIndex{3}$$:

Find the general term of a geometric sequence where $$a_{2} = −2$$ and $$a_{5}=\frac{2}{125}$$.

Solution

To determine a formula for the general term we need $$a_{1}$$ and $$r$$. A nonlinear system with these as variables can be formed using the given information and $$a_{n}=a_{1} r^{n-1} :$$:

$$\left\{\begin{array}{l}{a_{2}=a_{1} r^{2-1}} \\ {a_{5}=a_{1} r^{5-1}}\end{array}\right. \Longrightarrow \left\{\begin{array}{l}{-2=a_{1} r \quad\:\:\:\color{Cerulean}{Use\:a_{2}=-2.}} \\ {\frac{2}{125}=a_{1} r^{4} \quad\color{Cerulean}{Use\:a_{5}=\frac{2}{125}.}}\end{array}\right.$$

Solve for $$a_{1}$$ in the first equation,

$$-2=a_{1} r \quad \Rightarrow \quad \frac{-2}{r}=a_{1}$$
$$\frac{2}{125}=a_{1} r^{4}$$

Substitute $$a_{1} = \frac{-2}{r}$$ into the second equation and solve for $$r$$.

$$\frac{2}{125}=a_{1} r^{4}$$
$$\frac{2}{125}=\left(\frac{-2}{r}\right) r^{4}$$
$$\frac{2}{125}=-2 r^{3}$$
$$-\frac{1}{125}=r^{3}$$
$$-\frac{1}{5}=r$$

Back substitute to find $$a_{1}$$:

\begin{aligned} a_{1} &=\frac{-2}{r} \\ &=\frac{-2}{\left(-\frac{1}{5}\right)} \\ &=10 \end{aligned}

Therefore, $$a_{1} = 10$$ and $$r = −\frac{1}{5}$$.

$$a_{n}=10\left(-\frac{1}{5}\right)^{n-1}$$

Exercise $$\PageIndex{1}$$

Find an equation for the general term of the given geometric sequence and use it to calculate its $$6^{th}$$ term: $$2, \frac{4}{3},\frac{8}{9}, …$$

$$a_{n}=2\left(\frac{2}{3}\right)^{n-1} ; a_{6}=\frac{64}{243}$$

Geometric Series

A geometric series22 is the sum of the terms of a geometric sequence. For example, the sum of the first $$5$$ terms of the geometric sequence defined by $$a_{n}=3^{n+1}$$ follows:

\begin{aligned} S_{5} &=\sum_{n=1}^{5} 3^{n+1} \\ &=3^{1+1}+3^{2+1}+3^{3+1}+3^{4+1}+3^{5+1} \\ &=3^{2}+3^{3}+3^{4}+3^{5}+3^{6} \\ &=9+27+81+3^{5}+3^{6} \\ &=1,089 \end{aligned}

Adding $$5$$ positive integers is manageable. However, the task of adding a large number of terms is not. Therefore, we next develop a formula that can be used to calculate the sum of the first $$n$$ terms of any geometric sequence. In general,

$$S_{n}=a_{1}+a_{1} r+a_{1} r^{2}+\ldots+a_{1} r^{n-1}$$

Multiplying both sides by $$r$$ we can write,

$$r S_{n}=a_{1} r+a_{1} r^{2}+a_{1} r^{3}+\ldots+a_{1} r^{n}$$

Subtracting these two equations we then obtain,

$$S_{n}-r S_{n}=a_{1}-a_{1} r^{n}$$
$$S_{n}(1-r)=a_{1}\left(1-r^{n}\right)$$

Assuming $$r ≠ 1$$ dividing both sides by $$(1 − r)$$ leads us to the formula for the $$n$$th partial sum of a geometric sequence23:

$$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}(r \neq 1)$$

In other words, the $$n$$th partial sum of any geometric sequence can be calculated using the first term and the common ratio. For example, to calculate the sum of the first $$15$$ terms of the geometric sequence defined by $$a_{n}=3^{n+1}$$, use the formula with $$a_{1} = 9$$ and $$r = 3$$.

\begin{aligned} S_{15} &=\frac{a_{1}\left(1-r^{15}\right)}{1-r} \\ &=\frac{9 \cdot\left(1-3^{15}\right)}{1-3} \\ &=\frac{9(-14,348,906)}{-2} \\ &=64,570,077 \end{aligned}

Example $$\PageIndex{4}$$:

Find the sum of the first 10 terms of the given sequence: $$4, −8, 16, −32, 64,…$$

Solution

Determine whether or not there is a common ratio between the given terms.

$$r=\frac{-8}{4}=-2$$

Note that the ratio between any two successive terms is $$−2$$; hence, the given sequence is a geometric sequence. Use $$r = −2$$ and the fact that $$a_{1} = 4$$ to calculate the sum of the first $$10$$ terms,

\begin{aligned} S_{n} &=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \\ S_{10} &=\frac{\color{Cerulean}{4}\color{black}{\left[1-(\color{Cerulean}{-2}\color{black}{)}^{10}\right]}}{1-(\color{Cerulean}{-2}\color{black}{)}} ] \\ &=\frac{4(1-1,024)}{1+2} \\ &=\frac{4(-1,023)}{3} \\ &=-1,364 \end{aligned}

$$S_{10}=-1,364$$

Example $$\PageIndex{5}$$:

Evaluate: $$\sum_{n=1}^{6} 2(-5)^{n}$$.

Solution

In this case, we are asked to find the sum of the first $$6$$ terms of a geometric sequence with general term $$a_{n} = 2(−5)^{n}$$. Use this to determine the $$1^{st}$$ term and the common ratio $$r$$:

$$a_{1}=2(-5)^{1}=-10$$

To show that there is a common ratio we can use successive terms in general as follows:

\begin{aligned} r &=\frac{a_{n}}{a_{n-1}} \\ &=\frac{2(-5)^{n}}{2(-5)^{n-1}} \\ &=(-5)^{n-(n-1)} \\ &=(-5)^{1}\\&=-5 \end{aligned}

Use $$a_{1} = −10$$ and $$r = −5$$ to calculate the $$6^{th}$$ partial sum.

\begin{aligned} S_{n} &=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \\ S_{6} &=\frac{\color{Cerulean}{-10}\color{black}{\left[1-(\color{Cerulean}{-5}\color{black}{)}^{6}\right]}}{1-(\color{Cerulean}{-5}\color{black}{)}} \\ &=\frac{-10(1-15,625)}{1+5} \\ &=\frac{-10(-15,624)}{6} \\ &=26,040 \end{aligned}

$$26,040$$

Exercise $$\PageIndex{2}$$

Find the sum of the first 9 terms of the given sequence: $$-2,1,-1 / 2, \dots$$

$$S_{9}=-\frac{171}{128}$$

If the common ratio r of an infinite geometric sequence is a fraction where $$|r| < 1$$ (that is $$−1 < r < 1$$), then the factor $$(1 − r^{n})$$ found in the formula for the $$n$$th partial sum tends toward $$1$$ as $$n$$ increases. For example, if $$r = \frac{1}{10}$$ and $$n = 2, 4, 6$$ we have,

$$1-\left(\frac{1}{10}\right)^{2}=1-0.01=0.99$$
$$1-\left(\frac{1}{10}\right)^{4}=1-0.0001=0.9999$$
$$1-\left(\frac{1}{10}\right)^{6}=1-0.00001=0.999999$$

Here we can see that this factor gets closer and closer to 1 for increasingly larger values of $$n$$. This illustrates the idea of a limit, an important concept used extensively in higher-level mathematics, which is expressed using the following notation:

$$\lim _{n \rightarrow \infty}\left(1-r^{n}\right)=1$$ where $$|r|<1$$

This is read, “the limit of $$(1 − r^{n})$$ as $$n$$ approaches infinity equals $$1$$.” While this gives a preview of what is to come in your continuing study of mathematics, at this point we are concerned with developing a formula for special infinite geometric series. Consider the $$n$$th partial sum of any geometric sequence,

$$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}=\frac{a_{1}}{1-r}\left(1-r^{n}\right)$$

If $$|r| < 1$$ then the limit of the partial sums as n approaches infinity exists and we can write,

$$S_{n}=\frac{a_{1}}{1-r}\left(1-r^{n}\right)\quad\color{Cerulean}{\stackrel{\Longrightarrow}{n\rightarrow \infty }} \quad \color{black}{S_{\infty}}=\frac{a_{1}}{1-4}\cdot1$$

Therefore, a convergent geometric series24 is an infinite geometric series where $$|r| < 1$$; its sum can be calculated using the formula:

$$S_{\infty}=\frac{a_{1}}{1-r}$$

Example $$\PageIndex{6}$$:

Find the sum of the infinite geometric series: $$\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\dots$$

Solution

Determine the common ratio, Since the common ratio $$r = \frac{1}{3}$$ is a fraction between $$−1$$ and $$1$$, this is a convergent geometric series. Use the first term $$a_{1} = \frac{3}{2}$$ and the common ratio to calculate its sum

\begin{aligned} S_{\infty} &=\frac{a_{1}}{1-r} \\ &=\frac{\frac{3}{2}}{1-\left(\frac{1}{3}\right)} \\ &=\frac{\frac{3}{3}}{\frac{2}{3}} \\ &=\frac{3}{2} \cdot \frac{3}{2} \\ &=\frac{9}{4} \end{aligned}

$$S_{\infty}=\frac{9}{4}$$

Note

In the case of an infinite geometric series where $$|r| ≥ 1$$, the series diverges and we say that there is no sum. For example, if $$a_{n} = (5)^{n−1}$$ then $$r = 5$$ and we have

$$S_{\infty}=\sum_{n=1}^{\infty}(5)^{n-1}=1+5+25+\cdots$$

We can see that this sum grows without bound and has no sum.

Exercise $$\PageIndex{3}$$

Find the sum of the infinite geometric series: $$\sum_{n=1}^{\infty}-2\left(\frac{5}{9}\right)^{n-1}$$

$$-\frac{9}{2}$$

A repeating decimal can be written as an infinite geometric series whose common ratio is a power of $$1/10$$. Therefore, the formula for a convergent geometric series can be used to convert a repeating decimal into a fraction.

Example $$\PageIndex{7}$$:

Write as a fraction: $$1.181818…$$

Solution

Begin by identifying the repeating digits to the right of the decimal and rewrite it as a geometric progression.

\begin{aligned} 0.181818 \ldots &=0.18+0.0018+0.000018+\ldots \\ &=\frac{18}{100}+\frac{18}{10,000}+\frac{18}{1,000,000}+\ldots \end{aligned}

In this form we can determine the common ratio,

\begin{aligned} r &=\frac{\frac{18}{10,000}}{\frac{18}{100}} \\ &=\frac{18}{10,000} \times \frac{100}{18} \\ &=\frac{1}{100} \end{aligned}

Note that the ratio between any two successive terms is $$\frac{1}{100}$$. Use this and the fact that $$a_{1} = \frac{18}{100}$$ to calculate the infinite sum:

\begin{aligned} S_{\infty} &=\frac{a_{1}}{1-r} \\ &=\frac{\frac{18}{100}}{1-\left(\frac{1}{100}\right)} \\ &=\frac{\frac{18}{100}}{\frac{90}{100}} \\ &=\frac{18}{100} \cdot \frac{100}{99} \\ &=\frac{2}{11} \end{aligned}

Therefore, $$0.181818… = \frac{2}{11}$$ and we have,

$$1.181818 \ldots=1+\frac{2}{11}=1 \frac{2}{11}$$

$$1 \frac{2}{11}$$

Example $$\PageIndex{8}$$:

A certain ball bounces back to two-thirds of the height it fell from. If this ball is initially dropped from $$27$$ feet, approximate the total distance the ball travels.

Solution

We can calculate the height of each successive bounce:

$$\begin{array}{l}{27 \cdot \frac{2}{3}=18 \text { feet } \quad \color{Cerulean} { Height\: of\: the\: first\: bounce }} \\ {18 \cdot \frac{2}{3}=12 \text { feet}\quad\:\color{Cerulean}{ Height \:of\: the\: second\: bounce }} \\ {12 \cdot \frac{2}{3}=8 \text { feet } \quad\:\: \color{Cerulean} { Height\: of\: the\: third\: bounce }}\end{array}$$

The total distance that the ball travels is the sum of the distances the ball is falling and the distances the ball is rising. The distances the ball falls forms a geometric series,

$$27+18+12+\dots \quad\color{Cerulean}{Distance\:the\:ball\:is\:falling}$$

where $$a_{1} = 27$$ and $$r = \frac{2}{3}$$. Because $$r$$ is a fraction between $$−1$$ and $$1$$, this sum can be calculated as follows:

\begin{aligned} S_{\infty} &=\frac{a_{1}}{1-r} \\ &=\frac{27}{1-\frac{2}{3}} \\ &=\frac{27}{\frac{1}{3}} \\ &=81 \end{aligned}

Therefore, the ball is falling a total distance of $$81$$ feet. The distances the ball rises forms a geometric series,

$$18+12+8+\cdots \quad\color{Cerulean}{Distance\:the\:ball\:is\:rising}$$

where $$a_{1} = 18$$ and $$r = \frac{2}{3}$$. Calculate this sum in a similar manner:

\begin{aligned} S_{\infty} &=\frac{a_{1}}{1-r} \\ &=\frac{18}{1-\frac{2}{3}} \\ &=\frac{18}{\frac{1}{3}} \\ &=54 \end{aligned}

Therefore, the ball is rising a total distance of $$54$$ feet. Approximate the total distance traveled by adding the total rising and falling distances:

$$81+54=135$$ feet

$$135$$ feet

Key Takeaways

• A geometric sequence is a sequence where the ratio $$r$$ between successive terms is constant.
• The general term of a geometric sequence can be written in terms of its first term $$a_{1}$$, common ratio $$r$$, and index $$n$$ as follows: $$a_{n} = a_{1} r^{n−1}$$.
• A geometric series is the sum of the terms of a geometric sequence.
• The $$n$$th partial sum of a geometric sequence can be calculated using the first term $$a_{1}$$ and common ratio $$r$$ as follows: $$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$$.
• The infinite sum of a geometric sequence can be calculated if the common ratio is a fraction between $$−1$$ and $$1$$ (that is $$|r| < 1$$) as follows: $$S_{\infty}=\frac{a_{1}}{1-r}$$. If $$|r| ≥ 1$$, then no sum exists.

Exercise $$\PageIndex{4}$$

Write the first $$5$$ terms of the geometric sequence given its first term and common ratio. Find a formula for its general term.

1. $$a_{1}=1 ; r=5$$
2. $$a_{1}=1 ; r=3$$
3. $$a_{1}=2 ; r=3$$
4. $$a_{1}=5 ; r=4$$
5. $$a_{1}=2 ; r=-3$$
6. $$a_{1}=6 ; r=-2$$
7. $$a_{1}=3 ; r=\frac{2}{3}$$
8. $$a_{1}=6 ; r=\frac{1}{2}$$
9. $$a_{1}=1.2 ; r=0.6$$
10. $$a_{1}=-0.6 ; r=-3$$

1. $$1,5,25,125,625 ; a_{n}=5^{n-1}$$

3. $$2,6,18,54,162 ; a_{n}=2(3)^{n-1}$$

5. $$2,-6,18,-54,162 ; a_{n}=2(-3)^{n-1}$$

7. $$3,2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27} ; a_{n}=3\left(\frac{2}{3}\right)^{n-1}$$

9. $$1.2,0.72,0.432,0.2592,0.15552 ; a_{n}=1.2(0.6)^{n-1}$$

Exercise $$\PageIndex{5}$$

Given the geometric sequence, find a formula for the general term and use it to determine the $$5^{th}$$ term in the sequence.

1. $$7,28,112, \dots$$
2. $$-2,-10,-50, \dots$$
3. $$2, \frac{1}{2}, \frac{1}{8}, \ldots$$
4. $$1, \frac{2}{5}, \frac{4}{25}, \ldots$$
5. $$8,4,2, \dots$$
6. $$6,2, \frac{2}{3}, \ldots$$
7. $$-1, \frac{2}{3},-\frac{4}{9}, \ldots$$
8. $$2,-\frac{3}{2}, \frac{9}{8}, \ldots$$
9. $$\frac{1}{3},-2,12, \dots$$
10. $$\frac{2}{5},-2,10, \dots$$
11. $$-3.6,-4.32,-5.184, \dots$$
12. $$0.8,-2.08,5.408, \dots$$
13. Find the general term and use it to determine the $$20^{th}$$ term in the sequence: $$1, \frac{x}{2}, \frac{x^{2}}{4}, \ldots$$
14. Find the general term and use it to determine the $$20^{th}$$ term in the sequence: $$2,-6 x, 18 x^{2} \ldots$$
15. The number of cells in a culture of a certain bacteria doubles every $$4$$ hours. If $$200$$ cells are initially present, write a sequence that shows the population of cells after every $$n$$th $$4$$-hour period for one day. Write a formula that gives the number of cells after any $$4$$-hour period.
16. A certain ball bounces back at one-half of the height it fell from. If this ball is initially dropped from $$12$$ feet, find a formula that gives the height of the ball on the $$n$$th bounce and use it to find the height of the ball on the $$6^{th}$$ bounce.
17. Given a geometric sequence defined by the recurrence relation $$a_{n} = 4a_{n−1}$$ where $$a_{1} = 2$$ and $$n > 1$$, find an equation that gives the general term in terms of $$a_{1}$$ and the common ratio $$r$$.
18. Given the geometric sequence defined by the recurrence relation $$a_{n} = 6a_{n−1}$$ where $$a_{1} = \frac{1}{2}$$ and $$n > 1$$, find an equation that gives the general term in terms of $$a_{1}$$ and the common ratio $$r$$.

1. $$a_{n}=7(4)^{n-1}, a_{5}=1,792$$

3. $$a_{n}=2\left(\frac{1}{4}\right)^{n-1}, a_{5}=\frac{1}{128}$$

5. $$a_{n}=8\left(\frac{1}{2}\right)^{n-1}, a_{5}=\frac{1}{2}$$

7. $$a_{n}=-\left(-\frac{2}{3}\right)^{n-1}, a_{5}=-\frac{16}{81}$$

9. $$a_{n}=\frac{1}{3}(-6)^{n-1}, a_{5}=432$$

11. $$a_{n}=-3.6(1.2)^{n-1}, a_{5}=-7.46496$$

13. $$a_{n}=\left(\frac{x}{2}\right)^{n-1} ; a_{20}=\frac{x^{19}}{2^{19}}$$

15. $$400$$ cells; $$800$$ cells; $$1,600$$ cells; $$3,200$$ cells; $$6,400$$ cells; $$12,800$$ cells; $$p_{n} = 400(2)^{n−1}$$ cells

17. $$a_{n}=2(4)^{n-1}$$

Exercise $$\PageIndex{6}$$

Given the terms of a geometric sequence, find a formula for the general term.

1. $$a_{1}=-3$$ and $$a_{6}=-96$$
2. $$a_{1}=5$$ and $$a_{4}=-40$$
3. $$a_{1}=-2$$ and $$a_{8}=-\frac{1}{64}$$
4. $$a_{1}=\frac{3}{4}$$ and $$a_{4}=-\frac{1}{36}$$
5. $$a_{2}=18$$ and $$a_{5}=486$$
6. $$a_{2}=10$$ and $$a_{7}=320$$
7. $$a_{4}=-2$$ and $$a_{9}=64$$
8. $$a_{3}=-\frac{4}{3}$$ and $$a_{6}=\frac{32}{81}$$
9. $$a_{5}=153.6$$ and $$a_{8}=9,830.4$$
10. $$a_{4}=-2.4 \times 10^{-3}$$ and $$a_{9}=-7.68 \times 10^{-7}$$

1. $$a_{n}=-3(2)^{n-1}$$

3. $$a_{n}=-2\left(\frac{1}{2}\right)^{n-1}$$

5. $$a_{n}=6(3)^{n-1}$$

7. $$a_{n}=\frac{1}{4}(-2)^{n-1}$$

9. $$a_{n}=0.6(4)^{n-1}$$

Exercise $$\PageIndex{7}$$

Find all geometric means between the given terms.

1. $$a_{1}=2$$ and $$a_{4}=250$$
2. $$a_{1}=\frac{1}{3}$$ and $$a_{6}=-\frac{1}{96}$$
3. $$a_{2}=-20$$ and $$a_{5}=-20,000$$
4. $$a_{3}=49$$ and $$a_{6}=-16,807$$

1. $$10, 50$$

3. $$-200; -2,000$$

Exercise $$\PageIndex{8}$$

Calculate the indicated sum.

1. $$a_{n}=2^{n+1} ; S_{12}$$
2. $$a_{n}=(-2)^{n+1} ; S_{12}$$
3. $$a_{n}=\left(\frac{1}{2}\right)^{n} ; S_{7}$$
4. $$a_{n}=\left(\frac{2}{3}\right)^{n-1} ; S_{6}$$
5. $$a_{n}=5(-3)^{n-1} ; S_{5}$$
6. $$a_{n}=-7(-4)^{n} ; S_{5}$$
7. $$a_{n}=2\left(-\frac{1}{4}\right)^{n} ; S_{5}$$
8. $$a_{n}=\frac{1}{3}(2)^{n+1} ; S_{10}$$
9. $$\sum_{n=1}^{5} 5^{n}$$
10. $$\sum_{n=1}^{6}(-4)^{n}$$
11. $$\sum_{k=1}^{10} 2^{k+1}$$
12. $$\sum_{k=1}^{14} 2^{k-1}$$
13. $$\sum_{k=1}^{10}-2(3)^{k}$$
14. $$\sum_{k=1}^{8} 5(-2)^{k}$$
15. $$\sum_{n=1}^{5} 2\left(\frac{1}{2}\right)^{n+2}$$
16. $$\sum_{n=1}^{4}-3\left(\frac{2}{3}\right)^{n}$$
17. $$a_{n}=\left(\frac{1}{5}\right)^{n} ; S_{\infty}$$
18. $$a_{n}=\left(\frac{2}{3}\right)^{n-1} ; S_{\infty}$$
19. $$a_{n}=2\left(-\frac{3}{4}\right)^{n-1} ; S_{\infty}$$
20. $$a_{n}=3\left(-\frac{1}{6}\right)^{n} ; S_{\infty}$$
21. $$a_{n}=-2\left(\frac{1}{2}\right)^{n+1} ; S_{\infty}$$
22. $$a_{n}=-\frac{1}{3}\left(-\frac{1}{2}\right)^{n} ; S_{\infty}$$
23. $$\sum_{n=1}^{\infty} 2\left(\frac{1}{3}\right)^{n-1}$$
24. $$\sum_{n=1}^{\infty}\left(\frac{1}{5}\right)^{n}$$
25. $$\sum_{n=1}^{\infty} 3(2)^{n-2}$$
26. $$\sum_{n=1}^{\infty}-\frac{1}{4}(3)^{n-2}$$
27. $$\sum_{n=1}^{\infty} \frac{1}{2}\left(-\frac{1}{6}\right)^{n}$$
28. $$\sum_{n=1}^{\infty} \frac{1}{3}\left(-\frac{2}{5}\right)^{n}$$

1. $$16,380$$

3. $$\frac{127}{128}$$

5. $$305$$

7. $$−\frac{205}{512}$$

9. $$3,905$$

11. $$4,092$$

13. $$−177,144$$

15. $$\frac{31}{64}$$

17. $$\frac{1}{4}$$

19. $$\frac{8}{7}$$

21. $$−1$$

23. $$3$$

25. No sum

27. $$−\frac{1}{14}$$

Exercise $$\PageIndex{9}$$

Write as a mixed number.

1. $$1.222…$$
2. $$5.777 …$$
3. $$2.252525…$$
4. $$3.272727…$$
5. $$1.999…$$
6. $$1.090909…$$
7. Suppose you agreed to work for pennies a day for $$30$$ days. You will earn $$1$$ penny on the first day, $$2$$ pennies the second day, $$4$$ pennies the third day, and so on. How many total pennies will you have earned at the end of the $$30$$ day period? What is the dollar amount?
8. An initial roulette wager of $$$100$$ is placed (on red) and lost. To make up the difference, the player doubles the bet and places a$$$200$$ wager and loses. Again, to make up the difference, the player doubles the wager to \$$$400$$ and loses. If the player continues doubling his bet in this manner and loses $$7$$ times in a row, how much will he have lost in total?
9. A certain ball bounces back to one-half of the height it fell from. If this ball is initially dropped from $$12$$ feet, approximate the total distance the ball travels.
10. A golf ball bounces back off of a cement sidewalk three-quarters of the height it fell from. If the ball is initially dropped from $$8$$ meters, approximate the total distance the ball travels.
11. A structured settlement yields an amount in dollars each year, represented by $$n$$, according to the formula $$p_{n} = 6,000(0.80)^{n−1}$$. What is the total amount gained from the settlement after $$10$$ years?
12. Beginning with a square, where each side measures $$1$$ unit, inscribe another square by connecting the midpoints of each side. Continue inscribing squares in this manner indefinitely, as pictured:

Find the sum of the area of all squares in the figure. (Hint: Begin by finding the sequence formed using the areas of each square.)

1. $$1\frac{2}{9}$$

3. $$2\frac{25}{99}$$

5. $$2$$

7. $$1,073,741,823$$ pennies; $$\ 10,737,418.23$$

9. $$36$$ feet

11. $$\ 26,778.77$$

Exercise $$\PageIndex{10}$$

Categorize the sequence as arithmetic, geometric, or neither. Give the common difference or ratio, if it exists.

1. $$-12,24,-48, \dots$$
2. $$-7,-5,-3, \dots$$
3. $$-3,-11,-19, \dots$$
4. $$4,9,16, \dots$$
5. $$2, \frac{3}{2}, \frac{4}{3}, \dots$$
6. $$\frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \dots$$
7. $$\frac{1}{6},-\frac{1}{6},-\frac{1}{2}, \ldots$$
8. $$\frac{1}{3}, \frac{1}{4}, \frac{3}{16}, \dots$$
9. $$\frac{1}{2}, \frac{1}{4}, \frac{1}{6} \dots$$
10. $$-\frac{1}{10},-\frac{1}{5},-\frac{3}{10}, \dots$$
11. $$1.26,0.252,0.0504, \dots$$
12. $$0.02,0.08,0.18, \dots$$
13. $$1,-1,1,-1, \ldots$$
14. $$0,0,0, \dots$$

1. Geometric; $$r = −2$$

3. Arithmetic; $$d = −8$$

5. Neither

7. Arithmetic; $$d = −\frac{1}{3}$$

9. Neither

11. Geometric; $$r = 0.2$$

13. Geometric; $$r = −1$$

Exercise $$\PageIndex{11}$$

Categorize the sequence as arithmetic or geometric, and then calculate the indicated sum.

1. $$a_{n}=3(5)^{n-1} ; S_{8}$$
2. $$a_{n}=5-6 n ; S_{22}$$
3. $$a_{n}=2 n ; S_{14}$$
4. $$a_{n}=2^{n} ; S_{10}$$
5. $$a_{n}=-2\left(\frac{1}{7}\right)^{n-1} ; S_{\infty}$$
6. $$a_{n}=-2+\frac{1}{7} n ; S_{8}$$

1. Geometric; $$292,968$$

3. Arithmetic; $$210$$

5. Geometric; $$−\frac{7}{3}$$

Exercise $$\PageIndex{12}$$

Calculate the indicated sum.

1. $$\sum_{n=1}^{50}(3 n-5)$$
2. $$\sum_{n=1}^{25}(4-8 n)$$
3. $$\sum_{n=1}^{12}(-2)^{n-1}$$
4. $$\sum_{n=1}^{\infty} 5\left(-\frac{1}{2}\right)^{n-1}$$
5. $$\sum_{n=1}^{40} 5$$
6. $$\sum_{n=1}^{\infty} 0.6^{n}$$

1. $$3,575$$

3. $$−1,365$$

5. $$200$$

Exercise $$\PageIndex{13}$$

1. Use the techniques found in this section to explain why $$0.999… = 1$$.
2. Construct a geometric sequence where $$r = 1$$. Explore the $$n$$th partial sum of such a sequence. What conclusions can we make?

1. Answer may vary

Footnotes

18A sequence of numbers where each successive number is the product of the previous number and some constant $$r$$.

19Used when referring to a geometric sequence.

20The constant $$r$$ that is obtained from dividing any two successive terms of a geometric sequence; $$\frac{a_{n}}{a_{n-1}}=r$$.

21The terms between given terms of a geometric sequence.

22The sum of the terms of a geometric sequence.

23The sum of the first n terms of a geometric sequence, given by the formula: $$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} , r\neq 1$$.

24An infinite geometric series where $$|r| < 1$$ whose sum is given by the formula:$$S_{\infty}=\frac{a_{1}}{1-r}$$.