This is “Arithmetic Sequences and Series”, section 9.2 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.

For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. You may also download a PDF copy of this book (41 MB) or just this chapter (1 MB), suitable for printing or most e-readers, or a .zip file containing this book's HTML files (for use in a web browser offline).

Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

9.2 Arithmetic Sequences and Series

Learning Objectives

  1. Identify the common difference of an arithmetic sequence.
  2. Find a formula for the general term of an arithmetic sequence.
  3. Calculate the nth partial sum of an arithmetic sequence.

Arithmetic Sequences

An arithmetic sequenceA sequence of numbers where each successive number is the sum of the previous number and some constant d., or arithmetic progressionUsed when referring to an arithmetic sequence., is a sequence of numbers where each successive number is the sum of the previous number and some constant d.

an=an1+dArithmeticSequence

And because anan1=d, the constant d is called the common differenceThe constant d that is obtained from subtracting any two successive terms of an arithmetic sequence; anan1=d.. For example, the sequence of positive odd integers is an arithmetic sequence,

1,3,5,7,9,

Here a1=1 and the difference between any two successive terms is 2. We can construct the general term an=an1+2 where,

a1=1a2=a1+2=1+2=3a3=a2+2=3+2=5a4=a3+2=5+2=7a5=a4+2=7+2=9

In general, given the first term a1 of an arithmetic sequence and its common difference d, we can write the following:

a2=a1+da3=a2+d=(a1+d)+d=a1+2da4=a3+d=(a1+2d)+d=a1+3da5=a4+d=(a1+3d)+d=a1+4d

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

an=a1+(n1)dArithmeticSequence

In fact, any general term that is linear in n defines an arithmetic sequence.

Example 1

Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100th term: 7,10,13,16,19,

Solution:

Begin by finding the common difference,

d=107=3

Note that the difference between any two successive terms is 3. The sequence is indeed an arithmetic progression where a1=7 and d=3.

an=a1+(n1)d=7+(n1)3=7+3n3=3n+4

Therefore, we can write the general term an=3n+4. Take a minute to verify that this equation describes the given sequence. Use this equation to find the 100th term:

a100=3(100)+4=304

Answer: an=3n+4; a100=304

The common difference of an arithmetic sequence may be negative.

Example 2

Find an equation for the general term of the given arithmetic sequence and use it to calculate its 75th term: 6,4,2,0,2,

Solution:

Begin by finding the common difference,

d=46=2

Next find the formula for the general term, here a1=6 and d=2.

an=a1+(n1)d=6+(n1)(2)=62n+2=82n

Therefore, an=82n and the 75th term can be calculated as follows:

a75=82(75)=8150=142

Answer: an=82n; a100=142

The terms between given terms of an arithmetic sequence are called arithmetic meansThe terms between given terms of an arithmetic sequence..

Example 3

Find all terms in between a1=8 and a7=10 of an arithmetic sequence. In other words, find all arithmetic means between the 1st and 7th terms.

Solution:

Begin by finding the common difference d. In this case, we are given the first and seventh term:

an=a1+(n1)dUsen=7.a7=a1+(71)da7=a1+6d

Substitute a1=8 and a7=10 into the above equation and then solve for the common difference d.

10=8+6d18=6d3=d

Next, use the first term a1=8 and the common difference d=3 to find an equation for the nth term of the sequence.

an=8+(n1)3=8+3n3=11+3n

With an=3n11, where n is a positive integer, find the missing terms.

a1=3(1)11=311=8a2=3(2)11=611=5a3=3(3)11=911=2a4=3(4)11=1211=1a5=3(5)11=1511=4a6=3(6)11=1811=7}arithmeticmeansa7=3(7)11=2111=10

Answer: −5, −2, 1, 4, 7

In some cases, the first term of an arithmetic sequence may not be given.

Example 4

Find the general term of an arithmetic sequence where a3=1 and a10=48.

Solution:

To determine a formula for the general term we need a1 and d. A linear system with these as variables can be formed using the given information and an=a1+(n1)d:

{a3=a1+(31)da10=a1+(101)d{1=a1+2d48=a1+9dUsea3=1.Usea10=48.

Eliminate a1 by multiplying the first equation by −1 and add the result to the second equation.

{1=a1+2d48=a1+9d×(1)+{1=a12d48=a1+9d¯49=7d7=d

Substitute d=7 into 1=a1+2d to find a1.

1=a1+2(7)1=a1+1415=a1

Next, use the first term a1=15 and the common difference d=7 to find a formula for the general term.

an=a1+(n1)d=15+(n1)7=15+7n7=22+7n

Answer: an=7n22

Try this! Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100th term: 32,2,52,3,72,

Answer: an=12n+1; a100=51

Arithmetic Series

An arithmetic seriesThe sum of the terms of an arithmetic sequence. is the sum of the terms of an arithmetic sequence. For example, the sum of the first 5 terms of the sequence defined by an=2n1 follows:

S5=Σn=15(2n1)=[2(1)1]+[2(2)1]+[2(3)1]+[2(4)1]+[2(5)1]=1+3+5+7+9=25

Adding 5 positive odd integers, as we have done above, is managable. However, consider adding the first 100 positive odd integers. This would be very tedious. Therefore, we next develop a formula that can be used to calculate the sum of the first n terms, denoted Sn, of any arithmetic sequence. In general,

Sn=a1+(a1+d)+(a1+2d)++an

Writing this series in reverse we have,

Sn=an+(and)+(an2d)++a1

And adding these two equations together, the terms involving d add to zero and we obtain n factors of a1+an:

2Sn=(a1+an)+(a1+an)++(an+a1)2Sn=n(a1+an)

Dividing both sides by 2 leads us the formula for the nth partial sum of an arithmetic sequenceThe sum of the first n terms of an arithmetic sequence given by the formula: Sn=n(a1+an)2.:

Sn=n(a1+an)2

Use this formula to calculate the sum of the first 100 terms of the sequence defined by an=2n1. Here a1=1 and a100=199.

S100=100(a1+a100)2=100(1+199)2=10,000

Example 5

Find the sum of the first 50 terms of the given sequence: 4, 9, 14, 19, 24, …

Solution:

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

d=94=5

Note that the difference between any two successive terms is 5. The sequence is indeed an arithmetic progression and we can write

an=a1+(n1)d=4+(n1)5=4+5n5=5n1

Therefore, the general term is an=5n1. To calculate the 50th partial sum of this sequence we need the 1st and the 50th terms:

a1=4a50=5(50)1=249

Next use the formula to determine the 50th partial sum of the given arithmetic sequence.

Sn=n(a1+an)2S50=50.(a1+a50)2=50(4+249)2=25(253)=6,325

Answer: S50=6,325

Example 6

Evaluate: Σn=135(104n).

Solution:

In this case, we are asked to find the sum of the first 35 terms of an arithmetic sequence with general term an=104n. Use this to determine the 1st and the 35th term.

a1=104(1)=6a35=104(35)=130

Next use the formula to determine the 35th partial sum.

Sn=n(a1+an)2S35=35(a1+a35)2=35[6+(130)]2=35(124)2=2,170

Answer: −2,170

Example 7

The first row of seating in an outdoor amphitheater contains 26 seats, the second row contains 28 seats, the third row contains 30 seats, and so on. If there are 18 rows, what is the total seating capacity of the theater?

Figure 9.2

Roman Theater (Wikipedia)

Solution:

Begin by finding a formula that gives the number of seats in any row. Here the number of seats in each row forms a sequence:

26,28,30,

Note that the difference between any two successive terms is 2. The sequence is an arithmetic progression where a1=26 and d=2.

an=a1+(n1)d=26+(n1)2=26+2n2=2n+24

Therefore, the number of seats in each row is given by an=2n+24. To calculate the total seating capacity of the 18 rows we need to calculate the 18th partial sum. To do this we need the 1st and the 18th terms:

a1=26a18=2(18)+24=60

Use this to calculate the 18th partial sum as follows:

Sn=n(a1+an)2S18=18(a1+a18)2=18(26+60)2=9(86)=774

Answer: There are 774 seats total.

Try this! Find the sum of the first 60 terms of the given sequence: 5, 0, −5, −10, −15, …

Answer: S60=8,550

Key Takeaways

  • An arithmetic sequence is a sequence where the difference d between successive terms is constant.
  • The general term of an arithmetic sequence can be written in terms of its first term a1, common difference d, and index n as follows: an=a1+(n1)d.
  • An arithmetic series is the sum of the terms of an arithmetic sequence.
  • The nth partial sum of an arithmetic sequence can be calculated using the first and last terms as follows: Sn=n(a1+an)2.

Topic Exercises

    Part A: Arithmetic Sequences

      Write the first 5 terms of the arithmetic sequence given its first term and common difference. Find a formula for its general term.

    1. a1=5; d=3

    2. a1=12; d=2

    3. a1=15; d=5

    4. a1=7; d=4

    5. a1=12; d=1

    6. a1=23; d=13

    7. a1=1; d=12

    8. a1=54; d=14

    9. a1=1.8; d=0.6

    10. a1=4.3; d=2.1

      Given the arithmetic sequence, find a formula for the general term and use it to determine the 100th term.

    1. 3, 9, 15, 21, 27,…

    2. 3, 8, 13, 18, 23,…

    3. −3, −7, −11, −15, −19,…

    4. −6, −14, −22, −30, −38,…

    5. −5, −10, −15, −20, −25,…

    6. 2, 4, 6, 8, 10,…

    7. 12, 52, 92, 132, 172,…

    8. 13, 23, 53, 83, 113,…

    9. 13, 0, 13, 23, −1,…

    10. 14, 12, 54, −2, 114,…

    11. 0.8, 2, 3.2, 4.4, 5.6,…

    12. 4.4, 7.5, 10.6, 13.7, 16.8,…

    13. Find the 50th positive odd integer.

    14. Find the 50th positive even integer.

    15. Find the 40th term in the sequence that consists of every other positive odd integer: 1, 5, 9, 13,…

    16. Find the 40th term in the sequence that consists of every other positive even integer: 2, 6, 10, 14,…

    17. What number is the term 355 in the arithmetic sequence −15, −5, 5, 15, 25,…?

    18. What number is the term −172 in the arithmetic sequence 4, −4, −12, −20, −28,…?

    19. Given the arithmetic sequence defined by the recurrence relation an=an1+5 where a1=2 and n>1, find an equation that gives the general term in terms of a1 and the common difference d.

    20. Given the arithmetic sequence defined by the recurrence relation an=an19 where a1=4 and n>1, find an equation that gives the general term in terms of a1 and the common difference d.

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

    1. a1=6 and a7=42

    2. a1=12 and a12=6

    3. a1=19 and a26=56

    4. a1=9 and a31=141

    5. a1=16 and a10=376

    6. a1=54 and a11=654

    7. a3=6 and a26=40

    8. a3=16 and a15=76

    9. a4=8 and a23=30

    10. a5=7 and a37=135

    11. a4=2310 and a21=252

    12. a3=18 and a12=112

    13. a5=13.2 and a26=61.5

    14. a4=1.2 and a13=12.3

      Find all arithmetic means between the given terms.

    1. a1=3 and a6=17

    2. a1=5 and a5=7

    3. a2=4 and a8=7

    4. a5=12 and a9=72

    5. a5=15 and a7=21

    6. a6=4 and a11=1

    Part B: Arithmetic Series

      Calculate the indicated sum given the formula for the general term.

    1. an=3n+5; S100

    2. an=5n11; S100

    3. an=12n; S70

    4. an=132n; S120

    5. an=12n34; S20

    6. an=n35; S150

    7. an=455n; S65

    8. an=2n48; S95

    9. an=4.41.6n; S75

    10. an=6.5n3.3; S67

      Evaluate.

    1. n=1160(3n)
    2. n=1121(2n)
    3. n=1250(4n3)
    4. n=1120(2n+12)
    5. n=170(198n)
    6. n=1220(5n)
    7. n=160(5212n)
    8. n=151(38n+14)
    9. n=1120(1.5n2.6)
    10. n=1175(0.2n1.6)
    11. Find the sum of the first 200 positive integers.

    12. Find the sum of the first 400 positive integers.

      The general term for the sequence of positive odd integers is given by an=2n1 and the general term for the sequence of positive even integers is given by an=2n. Find the following.

    1. The sum of the first 50 positive odd integers.

    2. The sum of the first 200 positive odd integers.

    3. The sum of the first 50 positive even integers.

    4. The sum of the first 200 positive even integers.

    5. The sum of the first k positive odd integers.

    6. The sum of the first k positive even integers.

    7. The first row of seating in a small theater consists of 8 seats. Each row thereafter consists of 3 more seats than the previous row. If there are 12 rows, how many total seats are in the theater?

    8. The first row of seating in an outdoor amphitheater contains 42 seats, the second row contains 44 seats, the third row contains 46 seats, and so on. If there are 22 rows, what is the total seating capacity of the theater?

    9. If a triangular stack of bricks has 37 bricks on the bottom row, 34 bricks on the second row and so on with one brick on top. How many bricks are in the stack?

    10. Each successive row of a triangular stack of bricks has one less brick until there is only one brick on top. How many rows does the stack have if there are 210 total bricks?

    11. A 10-year salary contract offers $65,000 for the first year with a $3,200 increase each additional year. Determine the total salary obligation over the 10 year period.

    12. A clock tower strikes its bell the number of times indicated by the hour. At one o’clock it strikes once, at two o’clock it strikes twice and so on. How many times does the clock tower strike its bell in a day?

    Part C: Discussion Board

    1. Is the Fibonacci sequence an arithmetic sequence? Explain.

    2. Use the formula for the nth partial sum of an arithmetic sequence Sn=n(a1+an)2 and the formula for the general term an=a1+(n1)d to derive a new formula for the nth partial sum Sn=n2[2a1+(n1)d]. Under what circumstances would this formula be useful? Explain using an example of your own making.

    3. Discuss methods for calculating sums where the index does not start at 1. For example, Σn=1535(3n+4)=1,659.

    4. A famous story involves Carl Friedrich Gauss misbehaving at school. As punishment, his teacher assigned him the task of adding the first 100 integers. The legend is that young Gauss answered correctly within seconds. What is the answer and how do you think he was able to find the sum so quickly?

Answers

  1. 5, 8, 11, 14, 17; an=3n+2

  2. 15, 10, 5, 0, −5; an=205n

  3. 12, 32, 52, 72, 92; an=n12

  4. 1, 12, 0, 12, −1; an=3212n

  5. 1.8, 2.4, 3, 3.6, 4.2; an=0.6n+1.2

  6. an=6n3; a100=597

  7. an=14n; a100=399

  8. an=5n; a100=500

  9. an=2n32; a100=3972

  10. an=2313n; a100=983

  11. an=1.2n0.4; a100=119.6

  12. 99

  13. 157

  14. 38

  15. an=5n3

  16. an=6n

  17. an=3n22

  18. an=23n12

  19. an=122n

  20. an=2n16

  21. an=11035n

  22. an=2.3n+1.7

  23. 1, 5, 9, 13

  24. 92, 5, 112, 6, 132

  25. 18

  1. 15,650

  2. −2,450

  3. 90

  4. −7,800

  5. −4,230

  6. 38,640

  7. 124,750

  8. −18,550

  9. −765

  10. 10,578

  11. 20,100

  12. 2,500

  13. 2,550

  14. k2

  15. 294 seats

  16. 247 bricks

  17. $794,000

  1. Answer may vary

  2. Answer may vary