9.2: Arithmetic Sequences and Series
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Learning Objectives
- Identify the common difference of an arithmetic sequence.
- Find a formula for the general term of an arithmetic sequence.
- Calculate the nth partial sum of an arithmetic sequence.
Arithmetic Sequences
An arithmetic sequence12, or arithmetic progression13, is a sequence of numbers where each successive number is the sum of the previous number and some constant d.
an=an−1+dArithmeticsequence
And because an−an−1=d, the constant d is called the common difference14. 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=an−1+2 where,
a1=1
a2=a1+2=1+2=3
a3=a2+2=3+2=5
a4=a3+2=5+2=7
a5=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+(n−1)dArithmeticSequence
In fact, any general term that is linear in n defines an arithmetic sequence.
Example 9.2.1:
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=10−7=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+(n−1)d=7+(n−1)⋅3=7+3n−3=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 9.2.2:
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=4−6=−2
Next find the formula for the general term, here a1=6 and d=−2.
an=a1+(n−1)d=6+(n−1)⋅(−2)=6−2n+2=8−2n
Therefore, an=8−2n and the 75th term can be calculated as follows:
a75=8−2(75)=8−150=−142
Answer:
an=8−2n;a100=−142
The terms between given terms of an arithmetic sequence are called arithmetic means15.
Example 9.2.3:
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+(n−1)dUsen=7.a7=a1+(7−1)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+(n−1)⋅3=−8+3n−3=−11+3n
With an=3n−11, where n is a positive integer, find the missing terms.
a1=3(1)−11=3−11=−8a2=3(2)−11=6−11=−5a3=3(3)−11=9−11=−2a4=3(4)−11=12−11=1a5=3(5)−11=15−11=4a6=3(6)−11=18−11=7a7=3(7)−11=21−111=10}arithmeticmeans
Answer:
−5,−2,1,4,7
In some cases, the first term of an arithmetic sequence may not be given.
Example 9.2.4:
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+(n−1)d:
{a3=a1+(3−1)da10=a1+(10−1)d⇒{−1=a1+2dUsea3=−1.48=a1+9dUsea10=48.
Eliminate a1 by multiplying the first equation by −1 and add the result to the second equation.
{−1=a1+2d48=a1+9dx(−1)⟹+{1=−a1−2d48=a1+9d
49=7d7=d
Substitute d=7 into −1=a1+2d to find a1.
−1=a1+2(7)
−1=a1+14
−15=a1
Next, use the first term a1=−15 and the common difference d=7 to find a formula for the general term.
an=a1+(n−1)d=−15+(n−1)⋅7=−15+7n−7=−22+7n
Answer:
an=7n−22
Exercise 9.2.1
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
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Arithmetic Series
An arithmetic series16 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=2n−1 follows:
S5=5∑n=1(2n−1)=[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+(an−d)+(an−2d)+…+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 sequence17:
Sn=n(a1+an)2
Use this formula to calculate the sum of the first 100 terms of the sequence defined by an=2n−1. Here a1=1 and a100=199.
S100=100(a1+a100)2=100(1+199)2=10,000
Example 9.2.5:
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=9−4=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+(n−1)d=4+(n−1)⋅5=4+5n−5=5n−1
Therefore, the general term is an=5n−1. 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 9.2.6:
Evaluate: ∑35n=1(10−4n).
Solution
In this case, we are asked to find the sum of the first 35 terms of an arithmetic sequence with general term an=10−4n. Use this to determine the 1st and the 35th term.
a1=10−4(1)=6
a35=10−4(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 9.2.7:
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?

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+(n−1)d=26+(n−1)⋅2=26+2n−2=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.
Exercise 9.2.2
Find the sum of the first 60 terms of the given sequence: 5,0,−5,−10,−15,…
- Answer
-
S60=−8,550
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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+(n−1)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.
Exercise 9.2.3
Write the first 5 terms of the arithmetic sequence given its first term and common difference. Find a formula for its general term.
- a1=5;d=3
- a1=12;d=2
- a1=15;d=−5
- a1=7;d=−4
- a1=12;d=1
- a1=23;d=13
- a1=1;d=−12
- a1=−54;d=14
- a1=1.8;d=0.6
- a1=−4.3;d=2.1
- Answer
-
1. 5,8,11,14,17;an=3n+2
3. 15,10,5,0,−5;an=20−5n
5. 12,32,52,72,92;an=n−12
7. 1,12,0,−12,−1;an=32−12n
9. 1.8,2.4,3,3.6,4.2;an=0.6n+1.2
Exercise 9.2.4
Given the arithmetic sequence, find a formula for the general term and use it to determine the 100th term.
- 3,9,15,21,27,…
- 3,8,13,18,23,…
- −3,−7,−11,−15,−19,…
- −6,−14,−22,−30,−38,…
- −5,−10,−15,−20,−25,…
- 2,4,6,8,10,…
- 12,52,92,132,172,…
- −13,23,53,83,113,…
- 13,0,−13,−23,−1,…
- 14,−12,−54,−2,−114,…
- 0.8,2,3.2,4.4,5.6,…
- 4.4,7.5,10.6,13.7,16.8,…
- Find the 50th positive odd integer.
- Find the 50th positive even integer.
- Find the 40th term in the sequence that consists of every other positive odd integer: 1,5,9,13,…
- Find the 40th term in the sequence that consists of every other positive even integer: 2,6,10,14,…
- What number is the term 355 in the arithmetic sequence −15,−5,5,15,25,…?
- What number is the term −172 in the arithmetic sequence 4,−4,−12,−20,−28,…?
- Given the arithmetic sequence defined by the recurrence relation an=an−1+5 where a1=2 and n>1, find an equation that gives the general term in terms of a1 and the common difference d.
- Given the arithmetic sequence defined by the recurrence relation an=an−1−9 where a1=4 and n>1, find an equation that gives the general term in terms of a1 and the common difference d.
- Answer
-
1. an=6n−3;a100=597
3. an=1−4n;a100=−399
5. an=−5n;a100=−500
7. an=2n−32;a100=3972
9. an=23−13n;a100=−983
11. an=1.2n−0.4;a100=119.6
13. 99
15. 157
17. 38
19. an=5n−3
Exercise 9.2.5
Given the terms of an arithmetic sequence, find a formula for the general term.
- a1=6 and a7=42
- a1=−12 and a12=−6
- a1=−19 and a26=56
- a1=−9 and a31=141
- a1=16 and a10=376
- a1=54 and a11=654
- a3=6 and a26=−40
- a3=16 and a15=76
- a4=−8 and a23=30
- a5=−7 and a37=−135
- a4=−2310 and a21=−252
- a3=18 and a12=−112
- a5=13.2 and a26=61.5
- a4=−1.2 and a13=12.3
- Answer
-
1. an=6n
3. an=3n−22
5. an=23n−12
7. an=12−2n
9. an=2n−16
11. an=110−35n
13. an=2.3n+1.7
Exercise 9.2.6
Find all arithmetic means between the given terms.
- a1=−3 and a6=17
- a1=5 and a5=−7
- a2=4 and a8=7
- a5=12 and a9=−72
- a5=15 and a7=21
- a6=4 and a11=−1
- Answer
-
1. 1,5,9,13
3. 92,5,112,6,132
5. 18
Exercise 9.2.7
Calculate the indicated sum given the formula for the general term.
- an=3n+5;S100
- an=5n−11;S100
- an=12−n,S70
- an=1−32n;S120
- an=12n−34;S20
- an=n−35;S150
- an=45−5n;S65
- an=2n−48;S95
- an=4.4−1.6n;S75
- an=6.5n−3.3;S67
- Answer
-
1. 15,650
3. −2,450
5. 90
7. −7,800
9. −4,230
Exercise 9.2.8
Evaluate.
- ∑160n=1(3n)
- ∑121n=1(−2n)
- ∑250n=1(4n−3)
- ∑120n=1(2n+12)
- ∑70n=1(19−8n)
- ∑220n=1(5−n)
- ∑60n=1(52−12n)
- ∑51n=1(38n+14)
- ∑120n=1(1.5n−2.6)
- ∑175n=1(−0.2n−1.6)
- Find the sum of the first 200 positive integers.
- Find the sum of the first 400 positive integers.
- Answer
-
1. 38,640
3. 124,750
5. −18,550
7. −765
9. 10,578
11. 20,100
Exercise 9.2.9
The general term for the sequence of positive odd integers is given by an=2n−1 and the general term for the sequence of positive even integers is given by an=2n. Find the following.
- The sum of the first 50 positive odd integers.
- The sum of the first 200 positive odd integers.
- The sum of the first 50 positive even integers.
- The sum of the first 200 positive even integers.
- The sum of the first k positive odd integers.
- The sum of the first k positive even integers.
- 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?
- 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?
- 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?
- 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?
- 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.
- 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?
- Answer
-
1. 2,500
3. 2,550
5. k2
7. 294 seats
9. 247 bricks
11. $794,000
Exercise 9.2.10
- Is the Fibonacci sequence an arithmetic sequence? Explain.
- 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+(n−1)d to derive a new formula for the nth partial sum Sn=n2[2a1+(n−1)d]. Under what circumstances would this formula be useful? Explain using an example of your own making.
- Discuss methods for calculating sums where the index does not start at 1. For example, ∑35n=1(3n+4)=1,659.
- 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?
- Answer
-
1. Answer may vary
3. Answer may vary
Footnotes
12A sequence of numbers where each successive number is the sum of the previous number and some constant d.
13Used when referring to an arithmetic sequence.
14The constant d that is obtained from subtracting any two successive terms of an arithmetic sequence; an−an−1=d.
15The terms between given terms of an arithmetic sequence.
16The sum of the terms of an arithmetic sequence.
17The sum of the first n terms of an arithmetic sequence given by the formula: Sn=n(a1+an)2.