9.3: Geometric Sequences and Series

Learning Objectives

• 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:

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

Answer:

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

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.

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

Answer:

$-15,-45$

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 $r = −\frac{1}{5}$.

Answer:

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

Exercise $\PageIndex{1}$

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

Answer

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

www.youtube.com/v/IGPEl9vloLY

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,

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

Answer:

$S_{10}=-1,364$

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.

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

Answer:

$26,040$

Exercise $\PageIndex{2}$

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

Answer

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

www.youtube.com/v/v-t3P95rWe8

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,

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

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

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

Answer:

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

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: $\sum_{n=1}^{\infty}-2\left(\frac{5}{9}\right)^{n-1}$

Answer

$-\frac{9}{2}$

www.youtube.com/v/KxsPVUyle_A

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,

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

Answer:

$1 \frac{2}{11}$

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:

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

Figure $\PageIndex{1}$

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

Answer:

$135$ 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. $a_{1}=-0.6 ; r=-3$

Answer

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 $r$.

Answer

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 $a_{9}=-7.68 \times 10^{-7}$

Answer

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 $a_{6}=-16,807$

Answer

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. $\sum_{n=1}^{\infty} \frac{1}{3}\left(-\frac{2}{5}\right)^{n}$

Answer

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:

Figure $\PageIndex{2}$

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.)

Answer

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. $0,0,0, \dots$

Answer

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. $a_{n}=-2+\frac{1}{7} n ; S_{8}$

Answer

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. $\sum_{n=1}^{\infty} 0.6^{n}$

Answer

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?

Answer

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