###
Subsection Arithmetic Sequences

A charter tour bus service charges $50 plus $15 for each passenger. The cost of a tour is then a function of the number of passengers. Because the number of passengers, \(n\text{,}\) can only be a positive integer, the function is a sequence. If \(C_n\) represents the cost of a tour for passengers, then

\begin{align*}
C_1 \amp = 50 + 15(1)= 65\\
C_2 \amp = 50 + 15(2) = 80\\
C_3 \amp = 50 + 15(3) = 95
\end{align*}

and in general

\begin{equation*}
C_n = 50 + 15n
\end{equation*}

Each term of this sequence can be obtained from the previous one by adding 15. A sequence in which each term can be obtained from the previous term by adding a fixed amount is called an arithmetic sequence.

The fixed amount that we add to each term is the difference between two successive terms and is called the common difference. In the example above the common difference is 15. If we denote the first term of an arithmetic sequence by \(a\) and the common difference by \(d\text{,}\) then the sequence can be defined recursively by

\begin{align*}
a_1 \amp = a\\
a_{n+1} \amp = a_n + d
\end{align*}

####
Example 9.19.

Find the first four terms of an arithmetic sequence with first term 6 and common difference 3.

## Solution.

The first term is 6, so we have \(a_1 = 6\text{.}\) To find each subsequent term, we add 3 to the previous term:

\begin{align*}
a_2 \amp = a_1 + 3 = 6 + 3 = 9\\
a_3 \amp = a_2 + 3 = 9 + 3 = 12\\
a_4 \amp = a_3 + 3 = 12 + 3 = 15
\end{align*}

The first four terms are \(6, \) \(9, \) \(12, \) and \(15\text{.}\)

####
Checkpoint 9.20. Practice 1.

An arithmetic sequence defines a linear function of \(n\text{.}\) In the figure below, compare the graph of the linear function \(f(x) = 2x + 3\text{,}\) whose domain is the set of all real numbers, and the graph of the arithmetic sequence \(a_n = 2n + 3\text{,}\) whose domain is the set of positive integers.

####
Note 9.21.

Notice that the common difference, 2, of the sequence corresponds to the slope of the linear function.

####
Checkpoint 9.22. QuickCheck 1.

###
Subsection The General Term of an Arithmetic Sequence

We have found a recursive definition for an arithmetic sequence, but we can also find a non-recursive definition. That is, we can find a formula for the general term. Consider an arithmetic sequence with first term \(a\) and common difference \(d\text{.}\) The

\begin{equation*}
\begin{aligned}[t]
\text{first term is}~~~~~~ \amp a\\
\text{second term is}~~~~~~ \amp a + d\\
\text{third term is}~~~~~~ \amp a + d + d = a + 2d\\
\text{fourth term is}~~~~~~ \amp a + d + d + d = a + 3d\\
n^{\text{th}}~ \text{term is}~~~~~~ \amp a + d + d + \cdots + d = a + (n - 1)d
\end{aligned}
\end{equation*}

Thus, we have the following property.

#### Arithmetic Sequence.

The \(n^{\text{th}}\) term of an arithmetic sequence is

\begin{equation*}
a_n = a + (n-1)d
\end{equation*}

We can use this formula to find a particular term of an arithmetic sequence if we know the first term and the common difference.

####
Example 9.23.

Find the fourteenth term of the arithmetic sequence \(-6,~{-1},~4,~\cdots\)

## Solution.

First, we find the common difference by subtracting any term from its successor:

\begin{equation*}
d = -1 - (-6) = 5
\end{equation*}

Then we use the formula with \(n = 14\) :

\begin{equation*}
a_{14}= -6 + (14-1)(5)= 59
\end{equation*}

####
Checkpoint 9.24. Practice 2.

####
Example 9.25.

You deposit $8000 in an account that pays 5% simple annual interest.

Find a formula for the amount of money in the account after \(n\) years.

How much money will be in the account after 20 years?

## Solution.

Each year, 0.05(8000) or $400 interest is added to the account balance. Thus, the annual balances,

\(b_n\text{,}\) form an arithmetic sequence whose first term is 8400 and whose common difference is 400. Hence,

\begin{equation*}
b_n = 8400 + (n-1)400
\end{equation*}

We substitute

\(n=20\) into the formula for the general term found in part (a).

\begin{equation*}
b_{20} = 8400 + (19)400 = 16,000
\end{equation*}

After 20 years there will be $16,000 in the account.

####
Checkpoint 9.26. Practice 3.

###
Subsection Geometric Sequences

A national junior chess tournament starts with 1024 invited contestants. At the end of each round, the winners move on to the next level. Thus, after each round there are half as many contestants as before, and the number of remaining contestants is a function of the number of rounds completed. Because the number of rounds \(n\) is a positive integer, the function is a sequence.

If \(C_n\) represents the number of contestants after \(n\) rounds of competition, then

\begin{align*}
C_1 \amp = 1024 \left(\dfrac{1}{2}\right) = 512\\
C_2 \amp = 512 \left(\dfrac{1}{2}\right) = 1024 \left(\dfrac{1}{2}\right)^2 = 256\\
C_3 \amp = 256 \left(\dfrac{1}{2}\right) = 1024 \left(\dfrac{1}{2}\right)^3 = 128\
\end{align*}

In general

\begin{gather*}
C_n = 1024 \left(\dfrac{1}{2}\right)^n
\end{gather*}

Each term of the sequence defined above can be obtained from the previous one by multiplying by \(\dfrac{1}{2}\text{.}\) A sequence in which each term can be obtained from the previous term by multiplying by a fixed amount is called a geometric sequence.

The fixed amount we multiply each term by is the ratio of two successive terms and is called the common ratio. In the example above the common ratio is \(\dfrac{1}{2}\text{.}\) If we denote the first term of a geometric sequence by \(a\) and the common ratio by \(r\text{,}\) then the sequence can be defined recursively by

\begin{equation*}
\begin{aligned}[t]
a_1 \amp = a\\
a_{n+1} \amp = r a_n\\
\end{aligned}
\end{equation*}

####
Example 9.27.

Find the first four terms of a geometric sequence whose first term is 64 and whose common ratio is \(\dfrac{5}{4}\text{.}\)

## Solution.

The first term is 64, so we have \(a_1=64\text{.}\) To find each subsequent term, we multipy the previous term by \(\dfrac{5}{4}\text{.}\)

\begin{align*}
a_2 \amp = \left(\dfrac{5}{4}\right)a_1 = \left(\dfrac{5}{4}\right)(64) = 80\\
a_3 \amp = \left(\dfrac{5}{4}\right)a_2 = \left(\dfrac{5}{4}\right)(80) = 100\\
a_4 \amp = \left(\dfrac{5}{4}\right)a_3 = \left(\dfrac{5}{4}\right)(100) = 125
\end{align*}

The first four terms are 64, 80, 100, and 125.

####
Checkpoint 9.28. Practice 4.

A geometric sequence defines an exponential function of \(n\text{.}\) In the figure below, compare the graphs of the exponential function \(f(x) = 100(2)^{x-1}\text{,}\) whose domain is the set of real numbers, and the geometric sequence \(a_n = 100(2)^{n-1}\text{,}\) whose domain is the set of positive integers.

####
Note 9.29.

Notice that the common ratio of the geometric sequence corresponds to the base of the exponential function.

Recall that an arithmetic sequence defines a linear function whose slope corresponds to the common difference. Just as an exponential function grows much faster in the long run than a linear function, so does a geometric sequence grow much faster than an arithmetic sequence.

####
Checkpoint 9.30. QuickCheck 2.

####
Example 9.31.

Identify the following sequences as arithmetic, geometric, or neither.

\(\displaystyle 3,~5,~7,~\cdots\)

\(\displaystyle 3,~-6,~12,~\cdots\)

\(\displaystyle 3,~6,~10,~\cdots\)

\(\displaystyle 3,~1,~\dfrac{1}{3},\cdots\)

## Solution.

This sequence is arithmetic. Each term is obtained from the previous term by adding 2.

This sequence is geometric. Each term is obtained from the previous term by multiplying by \(-2\text{.}\)

This sequence is neither arithmetic or geometric.

This sequence is geometric. Each term is obtained from the previous term by multiplying by \(\dfrac{1}{3}\text{.}\)

####
Checkpoint 9.32. Practice 5.

###
Subsection The General Term of a Geometric Sequence

We can now find a non-recursive formula for the general term of a geometric sequence. If we denote the first term of the geometric sequence by \(a\text{,}\) then the

\begin{equation*}
\begin{aligned}[t]
\text{second term is}~~~~~~ \amp ar\\
\text{third term is}~~~~~~ \amp ar \cdot r = ar^2\\
\text{fourth term is}~~~~~~ \amp ar^2 \cdot r = ar^3\\
n^{\text{th}}~ \text{term is}~~~~~~ \amp ar^{n-1}\\
\end{aligned}
\end{equation*}

In general we have the following property.

#### Geometric Sequence.

The \(n^{\text{th}}\) term of a geometric sequence is

\begin{equation*}
a_n = ar^{n-1}
\end{equation*}

We can use the formula to find a particular term of a geometric sequence if we know the first term and the common ratio.

####
Example 9.33.

Find the ninth term of the geometric sequence \(-24,~12,~-6,~\cdots\text{.}\)

## Solution.

First, we find the common ratio by dividing any term by its predecessor.

\begin{equation*}
\dfrac{12}{-24}=\dfrac{-1}{2}
\end{equation*}

Then we use the formula with \(a = -24, ~r = \dfrac{-1}{2}\text{,}\) and \(n = 9\text{.}\)

\begin{equation*}
a_9 = -24(\dfrac{-1}{2})^8 = \dfrac{-3}{32}
\end{equation*}

####
Checkpoint 9.34. Practice 6.

####
Example 9.35.

You deposit $8000 in an account that pays 5% annual interest compounded annually.

Find a formula for the amount of money in the account after \(n\) years.

How much money will be in the account after 20 years?

## Solution.

Each year, the account balance is multiplied by 1.05. Thus, the annual balances

\(c_n\) form a geometric sequence whose first term is 8400 and whose common ratio is 1.05. Hence,

\begin{equation*}
c_n = 8400(1.05)^{n-1}
\end{equation*}

We substitute

\(n = 20\) into the formula for the general term found in part (a).

\begin{equation*}
c_{20} = 8400(1.05)^{19} = 21,226.38164
\end{equation*}

After 20 years, there will be $21,226.38 in the account.

####
Checkpoint 9.36. QuickCheck 3.

####
Checkpoint 9.37. Practice 7.

####
Example 9.38.

Find a nonrecursive definition for each sequence.

\(\displaystyle a_1=2,~~a_n = a_{n-1}+3\)

\(\displaystyle b_1=2,~~b_{n+1} = 3b_n\)

## Solution.

From the definition we see that the first few terms of the sequence are 2, 5, 8, and 11. Because each new term is found by adding 3 to the previous term, we see that this is an arithmetic sequence with a common difference of 3. This means that the sequence has the form

\begin{equation*}
a_n = 2 + (n-1)3~~~\text{or}~~~~a_n = 3n - 1
\end{equation*}

The first four terms of this sequence are 2, 6, 18, and 54. Because each new term is found by multiplying the previous term by 3, we see that this is a geometric sequence with a common ratio of 3. This means that the sequence has the form

\begin{equation*}
b_n = 3\cdot 2^{n-1}
\end{equation*}

####
Checkpoint 9.39. Practice 8.

###
Exercises Homework 9.2

#### Exercise Group.

For Problems 1–12, identify the sequence as arithmetic, geometric, or neither.

##### 1.

\(-2, -6, -18, -54, \cdots\)

##### 2.

\(-2, -6, -10, -14, \cdots\)

##### 3.

\(16, 8, 0, -8,\cdots\)

##### 4.

\(16, 8, 4, 2, \cdots\)

##### 5.

\(-1, 1, -1, 1, \cdots\)

##### 6.

\(1, 3, 6, 10, \cdots\)

##### 7.

\(1, 4, 9, 16, \cdots\)

##### 8.

\(5, -5, 5, -5, \cdots\)

##### 9.

\(27, 9, 3, 1, \cdots\)

##### 10.

\(\dfrac{-1}{3}, \dfrac{1}{3}, 1, \dfrac{5}{3}, \cdots\)

##### 11.

\(\dfrac{2}{3}, -1, \dfrac{3}{2}, \dfrac{-9}{4}, \cdots\)

##### 12.

\(2, -8, 32, -124, \cdots\)

#### Exercise Group.

For Problems 13–24, find the first four terms of the sequence.

##### 13.

\(a = 2, d = 4\)

##### 14.

\(a = 7, d = 3\)

##### 15.

\(a = \dfrac{1}{2}, d = \dfrac{1}{4}\)

##### 16.

\(a = \dfrac{2}{3}, d = \dfrac{1}{3}\)

##### 17.

\(a = 2.7, d = -0.8\)

##### 18.

\(a = 5.9, d = -1.3\)

##### 19.

\(a = 5, r = -2\)

##### 20.

\(a = -4, r = 3\)

##### 21.

\(a = 9, r = \dfrac{2}{3}\)

##### 22.

\(a = 25, r = \dfrac{4}{5}\)

##### 23.

\(a = 60, r = 0.4\)

##### 24.

\(a = 100, r = 0.3\)

#### Exercise Group.

For Problems 25–32, find the next three terms, and an expression for the general term.

##### 25.

\(3, 7, 11, \cdots\)

##### 26.

\(-10, -20, -30, \cdots\)

##### 27.

\(-1, -5, -9, \cdots\)

##### 28.

\(-6, -1, 4, \cdots\)

##### 29.

\(\dfrac{2}{3}, \dfrac{4}{3}, \dfrac{8}{3}, \cdots\)

##### 30.

\(6, 3, \dfrac{3}{2}, \cdots\)

##### 31.

\(4, -2, 1, \cdots\)

##### 32.

\(\dfrac{1}{2}, \dfrac{-3}{2}, \dfrac{9}{2}, \cdots\)

#### 33.

Find the twelfth term in the arithmetic sequence \(2, \dfrac{5}{2}, 3, \cdots\)

#### 34.

Find the tenth term in the arithmetic sequence \(\dfrac{3}{2}, 2, \dfrac{13}{4}, \cdots\)

#### 35.

Find the eighth term in the geometric sequence \(-3, \dfrac{3}{2}, \dfrac{-3}{4}, \cdots\)

#### 36.

Find the sixth term in the geometric sequence \(-5, 1, \dfrac{-1}{5}, \cdots\)

#### 37.

Find the first term of a geometric sequence with fifth term \(48\) and common ratio \(2\text{.}\)

#### 38.

Find the first term of a geometric sequence with fifth term \(1\) and common ratio \(\dfrac{1}{2}\text{.}\)

#### 39.

How many terms are in the sequence \(\dfrac{1}{8}, \dfrac{1}{4}, \dfrac{1}{2}, \cdots, 512\text{?}\)

#### 40.

How many terms are in the sequence \(\dfrac{27}{64}, \dfrac{9}{16}, \dfrac{3}{4}, \cdots, \dfrac{64}{27}\text{?}\)

#### Exercise Group.

For Problems 41–48, find a nonrecursive definition for the sequence.

##### 41.

\(s_1=3,~~s_n = s_{n-1} + 2\)

##### 42.

\(c_1=6,~~c_n = c_{n-1} - 4\)

##### 43.

\(x_1=0,~~x_{n+1} = x_n - 3\)

##### 44.

\(y_1=-1,~~y_{n+1} = y_n + 5\)

##### 45.

\(d_1=24,~~d_{n+1} = \dfrac{-1}{2} d_n\)

##### 46.

\(r_1=27,~~r_{n+1} = \dfrac{2}{3} r_n \)

##### 47.

\(w_1=1;~~w_n=2w_{n-1}\)

##### 48.

\(q_1=7;~~q_n=3q_{n-1}\)

#### Exercise Group.

For Problems 49–56,

Write a formula for the general term of the sequence.

Answer the question in the problem.

##### 49.

An outdoor theater has 30 seats in the first row, 32 seats in the second row, 34 seats in the third row, and so on, with two more seats in any one row than in the previous. Let \(s_n\) be the number of seats in the \(n^{\text{th}}\) row of the theater. How many seats are there in the fiftieth row?

##### 50.

Gwynn is training for a triathlon. In her first week of training, she bicycled a total of 20 miles, then increased to 24 miles in her second week, 28 miles in her third week, and so on, increasing her mileage by 4 miles per week. Let \(m_n\) be Gwynn’s mileage in her \(n^{\text{th}}\) week of training. How many miles will she bicycle in her eighteenth week of training?

##### 51.

The cost of drilling a well increases the deeper you drill. Mel’s Wells charges $50 for the first five feet, $55 for the second five feet, $60 for the third five feet, and so on. Let \(d_n\) be the cost of drilling the \(n^{\text{th}}\) five feet. How much does Mel charge to drill from a depth of 65 feet to a depth of 70 feet?

##### 52.

The cost of having the windows washed in a high-rise office building depends on the height of the window. We Do Windows charges $0.50 per square yard of glass below 10 feet high, $0.60 per square yard for windows between 10 and 20 feet above the ground, $0.70 per square yard for windows between 20 and 30 feet above the ground, and so on. Let \(w_n\) be the cost of washing windows on the \(n^{\text{th}}\) floor if each floor is 10 feet high. How much does We Do Windows charge to wash 100 square yards of windows on the twenty-third floor?

##### 53.

Valerie’s grandmother deposited $500 into a college fund for Valerie on the day she was born. The fund earns 5% interest compounded annually. Let \(V_n\) be the value of the deposit, including interest, \(n\) years later. How much will the deposit be worth on Valerie’s eighteenth birthday?

##### 54.

When he was 25 years old, Bruce won $2000 on the lottery and deposited the money into a retirement fund. The fund earns 6% interest compounded annually. Let \(R_n\) be the value of the deposit, including interest, \(n\) years later. How much will the deposit be worth when Bruce turns 65 years old?

##### 55.

One hundred kilograms of a toxic chemical was dumped illegally into a clean reservoir. A filter can remove 20% of the chemical still present each week (so that 80% of the previous amount remains.) Let \(c_n\) be the amount of the chemical remaining after \(n\) weeks. How much of the chemical will remain in the water after 20 weeks?

##### 56.

A heart patient is given 40 milliliters of a medication by injection. Each hour, 15% of the medicine still present is eliminated from the body (so that 85% of the previous amount remains.) Let \(m_n\) be the amount of medicine remaining in the patient’s body after \(n\) hours. How much of the medication is left in the patient’s body after 10 hours?