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Arithmetic Sequences

Module by: First Last. E-mail the author

Summary: In this section, you will:

  • Find the common difference for an arithmetic sequence.
  • Write terms of an arithmetic sequence.
  • Use a recursive formula for an arithmetic sequence.
  • Use an explicit formula for an arithmetic sequence.

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Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year.

As an example, consider a woman who starts a small contracting business. She purchases a new truck for $25,000. After five years, she estimates that she will be able to sell the truck for $8,000. The loss in value of the truck will therefore be $17,000, which is $3,400 per year for five years. The truck will be worth $21,600 after the first year; $18,200 after two years; $14,800 after three years; $11,400 after four years; and $8,000 at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value.

Finding Common Differences

The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is –3,400.

A sequence, {25000, 21600, 18200, 14800, 8000}, that shows the terms differ only by -3400.

The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term.

A sequence {3, 6, 9, 12, 15, ...} that shows the terms only differ by 3.

A General Note: Arithmetic Sequence:

An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If a 1 a 1 is the first term of an arithmetic sequence and dd is the common difference, the sequence will be:

{ a n }={ a 1 , a 1 +d, a 1 +2d, a 1 +3d,...} { a n }={ a 1 , a 1 +d, a 1 +2d, a 1 +3d,...}

Example 1

Problem 1

Finding Common Differences

Is each sequence arithmetic? If so, find the common difference.

  1. {1,2,4,8,16,...} {1,2,4,8,16,...}
  2. {3,1,5,9,13,...} {3,1,5,9,13,...}
Solution

Subtract each term from the subsequent term to determine whether a common difference exists.

  1. The sequence is not arithmetic because there is no common difference.

    Eqn1.jpg

  2. The sequence is arithmetic because there is a common difference. The common difference is 4.

    Eqn2.jpg

Analysis

The graph of each of these sequences is shown in Figure 1. We can see from the graphs that, although both sequences show growth, a a is not linear whereas b b is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line.

Figure 1
Two graphs of arithmetic sequences. Graph (a) grows exponentially while graph (b) grows linearly.

Q&A:

If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference?

No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference.

Try It:

Exercise 1

Is the given sequence arithmetic? If so, find the common difference.

{18, 16, 14, 12, 10,} {18, 16, 14, 12, 10,}
Solution

The sequence is arithmetic. The common difference is 2. 2.

Try It:

Exercise 2

Is the given sequence arithmetic? If so, find the common difference.

{1, 3, 6, 10, 15,} {1, 3, 6, 10, 15,}
Solution

The sequence is not arithmetic because 3163. 3163.

Writing Terms of Arithmetic Sequences

Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of n n and d d into formula below.

a n = a 1 +(n1)d a n = a 1 +(n1)d

How To:

Given the first term and the common difference of an arithmetic sequence, find the first several terms.

  1. Add the common difference to the first term to find the second term.
  2. Add the common difference to the second term to find the third term.
  3. Continue until all of the desired terms are identified.
  4. Write the terms separated by commas within brackets.

Example 2

Problem 1

Writing Terms of Arithmetic Sequences

Write the first five terms of the arithmetic sequence with a 1 =17 a 1 =17 and d=3 d=3 .

Solution

Adding 3 3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.

The first five terms are {17,14,11,8,5} {17,14,11,8,5}

Analysis

As expected, the graph of the sequence consists of points on a line as shown in Figure 2.

Figure 2
Graph of the arithmetic sequence. The points form a negative line.

Try It:

Exercise 3

List the first five terms of the arithmetic sequence with a 1 =1 a 1 =1 and d=5 d=5 .

Solution

{ 1 ,   6 ,   11 ,   16 ,   21 } { 1 ,   6 ,   11 ,   16 ,   21 }

How To:

Given any the first term and any other term in an arithmetic sequence, find a given term.

  1. Substitute the values given for a 1 , a n ,n a 1 , a n ,n into the formula a n = a 1 +(n1)d a n = a 1 +(n1)d to solve for d. d.
  2. Find a given term by substituting the appropriate values for a 1 ,n, a 1 ,n, and d d into the formula a n = a 1 +(n1)d. a n = a 1 +(n1)d.

Example 3

Problem 1

Writing Terms of Arithmetic Sequences

Given a 1 =8 a 1 =8 and a 4 =14 a 4 =14 , find a 5 a 5 .

Solution

The sequence can be written in terms of the initial term 8 and the common difference d d.

{ 8,8+d,8+2d,8+3d } { 8,8+d,8+2d,8+3d }

We know the fourth term equals 14; we know the fourth term has the form a 1 +3d=8+3d a 1 +3d=8+3d .

We can find the common difference d d.

a n = a 1 +(n1)d a 4 = a 1 +3d a 4 =8+3d Write the fourth term of the sequence in terms of  a 1  and d. 14=8+3d Substitute 14 for  a 4 .  d=2 Solve for the common difference. a n = a 1 +(n1)d a 4 = a 1 +3d a 4 =8+3d Write the fourth term of the sequence in terms of  a 1  and d. 14=8+3d Substitute 14 for  a 4 .  d=2 Solve for the common difference.

Find the fifth term by adding the common difference to the fourth term.

a 5 = a 4 +2=16 a 5 = a 4 +2=16
Analysis

Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation a n = a 1 +( n1 )d. a n = a 1 +( n1 )d.

Try It:

Exercise 4

Given a 3 =7 a 3 =7 and a 5 =17 a 5 =17 , find a 2 a 2 .

Solution

a 2 = 2 a 2 = 2

Using Recursive Formulas for Arithmetic Sequences

Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given.

a n = a n1 +d n2 a n = a n1 +d n2

A General Note: Recursive Formula for an Arithmetic Sequence:

The recursive formula for an arithmetic sequence with common difference d d is:

a n = a n1 +d n2 a n = a n1 +d n2
(9)

How To:

Given an arithmetic sequence, write its recursive formula.

  1. Subtract any term from the subsequent term to find the common difference.
  2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences.

Example 4

Problem 1

Writing a Recursive Formula for an Arithmetic Sequence

Write a recursive formula for the arithmetic sequence.

{18741526, …} {18741526, …}
Solution

The first term is given as −18 −18 . The common difference can be found by subtracting the first term from the second term.

d=−7(−18)=11 d=−7(−18)=11

Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.

a 1 =18 a n = a n1 +11, for n2 a 1 =18 a n = a n1 +11, for n2
Analysis

We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure 3. The growth pattern of the sequence shows the constant difference of 11 units.

Figure 3
Graph of the arithmetic sequence. The points form a positive line.

Q&A:

Do we have to subtract the first term from the second term to find the common difference?

No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference.

Try It:

Exercise 5

Write a recursive formula for the arithmetic sequence.

{25 37 49 61 } {25 37 49 61 }
Solution

a 1 = 25 a n = a n 1 + 12 ,  for  n 2 a 1 = 25 a n = a n 1 + 12 ,  for  n 2

Using Explicit Formulas for Arithmetic Sequences

We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.

a n = a 1 + d ( n 1 ) a n = a 1 + d ( n 1 )

To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.

A sequence, {200, 150, 100, 50, 0, ...}, that shows the terms differ only by -50.

The common difference is 50 50 , so the sequence represents a linear function with a slope of 50 50 . To find the y y-intercept, we subtract 50 50 from 200:200(50)=200+50=250 200:200(50)=200+50=250 . You can also find the y y-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure 4.

Figure 4
Graph of the arithmetic sequence. The points form a negative line.

Recall the slope-intercept form of a line is y=mx+b. y=mx+b. When dealing with sequences, we use a n a n in place of y y and n n in place of x. x. If we know the slope and vertical intercept of the function, we can substitute them for m m and b b in the slope-intercept form of a line. Substituting 50 50 for the slope and 250 250 for the vertical intercept, we get the following equation:

a n =50n+250 a n =50n+250

We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is a n =20050(n1) a n =20050(n1) , which simplifies to a n =50n+250. a n =50n+250.

A General Note: Explicit Formula for an Arithmetic Sequence:

An explicit formula for the nth nth term of an arithmetic sequence is given by

a n = a 1 +d(n1) a n = a 1 +d(n1)
(16)

How To:

Given the first several terms for an arithmetic sequence, write an explicit formula.

  1. Find the common difference, a 2 a 1 . a 2 a 1 .
  2. Substitute the common difference and the first term into a n = a 1 +d(n1). a n = a 1 +d(n1).

Example 5

Problem 1

Writing the nth Term Explicit Formula for an Arithmetic Sequence

Write an explicit formula for the arithmetic sequence.

{212223242, …} {212223242, …}
Solution

The common difference can be found by subtracting the first term from the second term.

d = a 2 a 1 =122 =10 d = a 2 a 1 =122 =10

The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.

a n =2+10(n1) a n =10n8 a n =2+10(n1) a n =10n8
Analysis

The graph of this sequence, represented in Figure 5, shows a slope of 10 and a vertical intercept of 8 8 .

Figure 5
Graph of the arithmetic sequence. The points form a positive line.

Try It:

Exercise 6

Write an explicit formula for the following arithmetic sequence.

{50,47,44,41,} {50,47,44,41,}
Solution

a n = 53 3 n a n = 53 3 n

Finding the Number of Terms in a Finite Arithmetic Sequence

Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence.

How To:

Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms.

  1. Find the common difference d. d.
  2. Substitute the common difference and the first term into a n = a 1 +d(n1). a n = a 1 +d(n1).
  3. Substitute the last term for a n a n and solve for n. n.

Example 6

Problem 1
Finding the Number of Terms in a Finite Arithmetic Sequence

Find the number of terms in the finite arithmetic sequence.

{81–6...–41} {81–6...–41}
Solution

The common difference can be found by subtracting the first term from the second term.

18=7 18=7

The common difference is 7 7 . Substitute the common difference and the initial term of the sequence into the nth nth term formula and simplify.

a n = a 1 + d ( n 1 ) a n = 8 + 7 ( n 1 ) a n = 15 7 n a n = a 1 + d ( n 1 ) a n = 8 + 7 ( n 1 ) a n = 15 7 n

Substitute 41 41 for a n a n and solve for n n

41=157n 8=n 41=157n 8=n

There are eight terms in the sequence.

Try It:

Exercise 7

Find the number of terms in the finite arithmetic sequence.

{61116...56} {61116...56}
Solution

There are 11 terms in the sequence.

Solving Application Problems with Arithmetic Sequences

In many application problems, it often makes sense to use an initial term of a 0 a 0 instead of a 1 . a 1 . In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula:

a n = a 0 + d n a n = a 0 + d n

Example 7

Problem 1
Solving Application Problems with Arithmetic Sequences

A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week.

  1. Write a formula for the child’s weekly allowance in a given year.
  2. What will the child’s allowance be when he is 16 years old?
Solution
  1. The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2.

    Let A A be the amount of the allowance and n n be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get:

    A n = 1 + 2 n A n = 1 + 2 n
  2. We can find the number of years since age 5 by subtracting.

    16 5 = 11 16 5 = 11

    We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16.

    A 11 = 1 + 2 ( 11 ) = 23 A 11 = 1 + 2 ( 11 ) = 23

    The child’s allowance at age 16 will be $23 per week.

Try It:

Exercise 8

A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today?

Solution

The formula is T n =10+4n, T n =10+4n, and it will take her 42 minutes.

Media:

Access this online resource for additional instruction and practice with arithmetic sequences.

Key Equations

Table 1
recursive formula for nth term of an arithmetic sequence a n = a n1 +d n2 a n = a n1 +d n2
explicit formula for nth term of an arithmetic sequence a n = a 1 +d(n1) a n = a 1 +d(n1)

Key Concepts

  • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
  • The constant between two consecutive terms is called the common difference.
  • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example 1.
  • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example 2 and Example 3.
  • A recursive formula for an arithmetic sequence with common difference d d is given by a n = a n1 +d,n2. a n = a n1 +d,n2. See Example 4.
  • As with any recursive formula, the initial term of the sequence must be given.
  • An explicit formula for an arithmetic sequence with common difference d d is given by a n = a 1 +d(n1). a n = a 1 +d(n1). See Example 5.
  • An explicit formula can be used to find the number of terms in a sequence. See Example 6.
  • In application problems, we sometimes alter the explicit formula slightly to a n = a 0 +dn. a n = a 0 +dn. See Example 7.

Section Exercises

Verbal

Exercise 9

What is an arithmetic sequence?

Solution

A sequence where each successive term of the sequence increases (or decreases) by a constant value.

Exercise 10

How is the common difference of an arithmetic sequence found?

Exercise 11

How do we determine whether a sequence is arithmetic?

Solution

We find whether the difference between all consecutive terms is the same. This is the same as saying that the sequence has a common difference.

Exercise 12

What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence?

Exercise 13

Describe how linear functions and arithmetic sequences are similar. How are they different?

Solution

Both arithmetic sequences and linear functions have a constant rate of change. They are different because their domains are not the same; linear functions are defined for all real numbers, and arithmetic sequences are defined for natural numbers or a subset of the natural numbers.

Algebraic

For the following exercises, find the common difference for the arithmetic sequence provided.

Exercise 14

{ 5 , 11 , 17 , 23 , 29 , ... } { 5 , 11 , 17 , 23 , 29 , ... }

Exercise 15

{ 0 , 1 2 , 1 , 3 2 , 2 , ... } { 0 , 1 2 , 1 , 3 2 , 2 , ... }

Solution

The common difference is 1 2 1 2

For the following exercises, determine whether the sequence is arithmetic. If so find the common difference.

Exercise 16

{ 11.4 , 9.3 , 7.2 , 5.1 , 3 , ... } { 11.4 , 9.3 , 7.2 , 5.1 , 3 , ... }

Exercise 17

{ 4 , 16 , 64 , 256 , 1024 , ... } { 4 , 16 , 64 , 256 , 1024 , ... }

Solution

The sequence is not arithmetic because 1646416. 1646416.

For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference.

Exercise 18

a 1 =−25 a 1 =−25 , d=−9 d=−9

Exercise 19

a 1 =0 a 1 =0 , d= 2 3 d= 2 3

Solution

0, 2 3 , 4 3 ,2, 8 3 0, 2 3 , 4 3 ,2, 8 3

For the following exercises, write the first five terms of the arithmetic series given two terms.

Exercise 20

a 1 =17, a 7 =31 a 1 =17, a 7 =31

Exercise 21

a 13 =60, a 33 =160 a 13 =60, a 33 =160

Solution

0 , 5 , 10 , 15 , 20 0 , 5 , 10 , 15 , 20

For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference.

Exercise 22

First term is 3, common difference is 4, find the 5th term.

Exercise 23

First term is 4, common difference is 5, find the 4th term.

Solution

a 4 =19 a 4 =19

Exercise 24

First term is 5, common difference is 6, find the 8th term.

Exercise 25

First term is 6, common difference is 7, find the 6th term.

Solution

a 6 =41 a 6 =41

Exercise 26

First term is 7, common difference is 8, find the 7th term.

For the following exercises, find the first term given two terms from an arithmetic sequence.

Exercise 27

Find the first term or a 1 a 1 of an arithmetic sequence if a 6 =12 a 6 =12 and a 14 =28. a 14 =28.

Solution

a 1 =2 a 1 =2

Exercise 28

Find the first term or a 1 a 1 of an arithmetic sequence if a 7 =21 a 7 =21 and a 15 =42. a 15 =42.

Exercise 29

Find the first term or a 1 a 1 of an arithmetic sequence if a 8 =40 a 8 =40 and a 23 =115. a 23 =115.

Solution

a 1 =5 a 1 =5

Exercise 30

Find the first term or a 1 a 1 of an arithmetic sequence if a 9 =54 a 9 =54 and a 17 =102. a 17 =102.

Exercise 31

Find the first term or a 1 a 1 of an arithmetic sequence if a 11 =11 a 11 =11 and a 21 =16. a 21 =16.

Solution

a 1 =6 a 1 =6

For the following exercises, find the specified term given two terms from an arithmetic sequence.

Exercise 32

a 1 =33 a 1 =33 and a 7 =15. a 7 =15. Find a 4 . a 4 .

Exercise 33

a 3 =17.1 a 3 =17.1 and a 10 =15.7. a 10 =15.7. Find a 21 . a 21 .

Solution

a 21 =13.5 a 21 =13.5

For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence.

Exercise 34

a 1 =39;  a n = a n1 3 a 1 =39;  a n = a n1 3

Exercise 35

a 1 =19;  a n = a n1 1.4 a 1 =19;  a n = a n1 1.4

Solution

19,20.4,21.8,23.2,24.6 19,20.4,21.8,23.2,24.6

For the following exercises, write a recursive formula for each arithmetic sequence.

Exercise 36

a n ={ 40,60,80,... } a n ={ 40,60,80,... }

Exercise 37

a n ={17,26,35,...} a n ={17,26,35,...}

Solution

a 1 =17;  a n = a n1 +9 n2 a 1 =17;  a n = a n1 +9 n2

Exercise 38

a n ={1,2,5,...} a n ={1,2,5,...}

Exercise 39

a n ={12,17,22,...} a n ={12,17,22,...}

Solution

a 1 =12;  a n = a n1 +5 n2 a 1 =12;  a n = a n1 +5 n2

Exercise 40

a n ={15,7,1,...} a n ={15,7,1,...}

Exercise 41

a n ={8.9,10.3,11.7,...} a n ={8.9,10.3,11.7,...}

Solution

a 1 =8.9;  a n = a n1 +1.4 n2 a 1 =8.9;  a n = a n1 +1.4 n2

Exercise 42

a n ={0.52,1.02,1.52,...} a n ={0.52,1.02,1.52,...}

Exercise 43

a n ={ 1 5 , 9 20 , 7 10 ,... } a n ={ 1 5 , 9 20 , 7 10 ,... }

Solution

a 1 = 1 5 ;  a n = a n1 + 1 4 n2 a 1 = 1 5 ;  a n = a n1 + 1 4 n2

Exercise 44

a n ={ 1 2 , 5 4 ,2,... } a n ={ 1 2 , 5 4 ,2,... }

Exercise 45

a n ={ 1 6 , 11 12 ,2,... } a n ={ 1 6 , 11 12 ,2,... }

Solution

1 = 1 6 ;  a n = a n1 13 12 n2 1 = 1 6 ;  a n = a n1 13 12 n2

For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term.

Exercise 46

a n ={741...}; a n ={741...}; Find the 17th term.

Exercise 47

a n ={41118...}; a n ={41118...}; Find the 14th term.

Solution

a 1 =4;  a n = a n1 +7;  a 14 =95 a 1 =4;  a n = a n1 +7;  a 14 =95

Exercise 48

a n ={2610...}; a n ={2610...}; Find the 12th term.

For the following exercises, use the explicit formula to write the first five terms of the arithmetic sequence.

Exercise 49

a n =244n a n =244n

Solution

First five terms: 20,16,12,8,4. 20,16,12,8,4.

Exercise 50

a n = 1 2 n 1 2 a n = 1 2 n 1 2

For the following exercises, write an explicit formula for each arithmetic sequence.

Exercise 51

a n ={3,5,7,...} a n ={3,5,7,...}

Solution

a n =1+2n a n =1+2n

Exercise 52

a n ={32,24,16,...} a n ={32,24,16,...}

Exercise 53

a n ={595195...} a n ={595195...}

Solution

a n =105+100n a n =105+100n

Exercise 54

a n ={−17−217−417,...} a n ={−17−217−417,...}

Exercise 55

a n ={1.83.65.4...} a n ={1.83.65.4...}

Solution

a n =1.8n a n =1.8n

Exercise 56

a n ={−18.1,−16.2,−14.3,...} a n ={−18.1,−16.2,−14.3,...}

Exercise 57

a n ={15.8,18.5,21.2,...} a n ={15.8,18.5,21.2,...}

Solution

a n =13.1+2.7n a n =13.1+2.7n

Exercise 58

a n ={ 1 3 , 4 3 ,−3... } a n ={ 1 3 , 4 3 ,−3... }

Exercise 59

a n ={ 0, 1 3 , 2 3 ,... } a n ={ 0, 1 3 , 2 3 ,... }

Solution

a n = 1 3 n 1 3 a n = 1 3 n 1 3

Exercise 60

a n ={ 5, 10 3 , 5 3 , } a n ={ 5, 10 3 , 5 3 , }

For the following exercises, find the number of terms in the given finite arithmetic sequence.

Exercise 61

a n ={3,4,11...,60} a n ={3,4,11...,60}

Solution

There are 10 terms in the sequence.

Exercise 62

a n ={1.2,1.4,1.6,...,3.8} a n ={1.2,1.4,1.6,...,3.8}

Exercise 63

a n ={ 1 2 ,2, 7 2 ,...,8 } a n ={ 1 2 ,2, 7 2 ,...,8 }

Solution

There are 6 terms in the sequence.

Graphical

For the following exercises, determine whether the graph shown represents an arithmetic sequence.

Exercise 64

Graph of a scattered plot with labeled points: (1, -4), (2, -2), (3, 0), (4, 2), and (5, 4). The x-axis is labeled n and the y-axis is labeled a_n.

Exercise 65

Graph of a scattered plot with labeled points: (1, 1.5), (2, 2.25), (3, 3.375), (4, 5.0625), and (5, 7.5938). The x-axis is labeled n and the y-axis is labeled a_n.
Solution

The graph does not represent an arithmetic sequence.

For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence.

Exercise 66

a 1 =0,d=4 a 1 =0,d=4

Exercise 67

a 1 =9; a n = a n1 10 a 1 =9; a n = a n1 10

Solution
Graph of a scattered plot with labeled points: (1, 9), (2, -1), (3, -11), (4, -21), and (5, -31). The x-axis is labeled n and the y-axis is labeled a_n.

Exercise 68

a n =12+5n a n =12+5n

Technology

For the following exercises, follow the steps to work with the arithmetic sequence a n =3n2 a n =3n2 using a graphing calculator:

  • Press [MODE]
    • Select SEQ in the fourth line
    • Select DOT in the fifth line
    • Press [ENTER]
  • Press [Y=]
    • nMin nMin is the first counting number for the sequence. Set nMin=1 nMin=1
    • u(n) u(n) is the pattern for the sequence. Set u(n)=3n2 u(n)=3n2
    • u(nMin) u(nMin) is the first number in the sequence. Set u(nMin)=1 u(nMin)=1
  • Press [2ND] then [WINDOW] to go to TBLSET
    • Set TblStart=1 TblStart=1
    • Set ΔTbl=1 ΔTbl=1
    • Set Indpnt: Auto and Depend: Auto
  • Press [2ND] then [GRAPH] to go to the TABLE

Exercise 69

What are the first seven terms shown in the column with the heading u(n)? u(n)?

Solution

1,4,7,10,13,16,19 1,4,7,10,13,16,19

Exercise 70

Use the scroll-down arrow to scroll to n=50. n=50. What value is given for u(n)? u(n)?

Exercise 71

Press [WINDOW]. Set nMin=1,nMax=5,xMin=0,xMax=6,yMin=1, nMin=1,nMax=5,xMin=0,xMax=6,yMin=1, and yMax=14. yMax=14. Then press [GRAPH]. Graph the sequence as it appears on the graphing calculator.

Solution
Graph of a scattered plot with labeled points: (1, 1), (2, 4), (3, 7), (4, 10), and (5, 13). The x-axis is labeled n and the y-axis is labeled a_n.

For the following exercises, follow the steps given above to work with the arithmetic sequence a n = 1 2 n+5 a n = 1 2 n+5 using a graphing calculator.

Exercise 72

What are the first seven terms shown in the column with the heading u(n) u(n) in the TABLE feature?

Exercise 73

Graph the sequence as it appears on the graphing calculator. Be sure to adjust the WINDOW settings as needed.

Solution
Graph of a scattered plot with labeled points: (1, 5.5), (2, 6), (3, 6.5), (4, 7), and (5, 7.5). The x-axis is labeled n and the y-axis is labeled a_n.

Extensions

Exercise 74

Give two examples of arithmetic sequences whose 4th terms are 9. 9.

Exercise 75

Give two examples of arithmetic sequences whose 10th terms are 206. 206.

Solution

Answers will vary. Examples: a n =20.6n a n =20.6n and a n =2+20.4n. a n =2+20.4n.

Exercise 76

Find the 5th term of the arithmetic sequence {9b,5b,b,}. {9b,5b,b,}.

Exercise 77

Find the 11th term of the arithmetic sequence {3a2b,a+2b,a+6b}. {3a2b,a+2b,a+6b}.

Solution

a 11 =17a+38b a 11 =17a+38b

Exercise 78

At which term does the sequence {5.4,14.5,23.6,...} {5.4,14.5,23.6,...} exceed 151?

Exercise 79

At which term does the sequence { 17 3 , 31 6 , 14 3 ,... } { 17 3 , 31 6 , 14 3 ,... } begin to have negative values?

Solution

The sequence begins to have negative values at the 13th term, a 13 = 1 3 a 13 = 1 3

Exercise 80

For which terms does the finite arithmetic sequence { 5 2 , 19 8 , 9 4 ,..., 1 8 } { 5 2 , 19 8 , 9 4 ,..., 1 8 } have integer values?

Exercise 81

Write an arithmetic sequence using a recursive formula. Show the first 4 terms, and then find the 31st term.

Solution

Answers will vary. Check to see that the sequence is arithmetic. Example: Recursive formula: a 1 =3, a n = a n1 3. a 1 =3, a n = a n1 3. First 4 terms: 3,0,3,6 a 31 =87 3,0,3,6 a 31 =87

Exercise 82

Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28th term.

Glossary

arithmetic sequence:
a sequence in which the difference between any two consecutive terms is a constant
common difference:
the difference between any two consecutive terms in an arithmetic sequence

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