Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence. To add the first n terms, average the first and last terms and multiply by the number of terms: S_n = \frac{n(a_1 + a_n)}{2}. If you don’t know the last term, use S_n = \frac{n}{2}[2a_1 + (n - 1)d].

Before this lesson: You should know Arithmetic Sequences and Sigma Notation.

The Pairing Trick

Suppose you want 1 + 2 + 3 + \cdots + 100. Pair the first term with the last, the second with the second to last, and so on. Each pair adds to the same total: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101.

There are 100 terms, so there are 50 pairs. The sum is 50 \cdot 101 = 5050. This works for every arithmetic series, because the terms go up by the same amount from one end and down by that amount from the other.

Arithmetic Series Formulas

For an arithmetic sequence with first term a_1, common difference d and n terms:

S_n = \frac{n(a_1 + a_n)}{2}

S_n = \frac{n}{2}\left[2a_1 + (n - 1)d\right]

The second formula comes from the first. Replace a_n with the nth term formula a_1 + (n - 1)d.

1. Find a_1, d and n.
2. If you need it, find the last term with a_n = a_1 + (n - 1)d.
3. Substitute into a sum formula.
4. Simplify. Work inside the parentheses or brackets first.

Worked Examples

Example 1: Add 1 through 100

Find 1 + 2 + 3 + \cdots + 100.

Identify the parts: a_1 = 1, a_n = 100, n = 100

S_{100} = \frac{100(1 + 100)}{2}

S_{100} = \frac{100 \cdot 101}{2}

S_{100} = \frac{10{,}100}{2}

S_{100} = 5050

Answer: 5050

Example 2: Find the sum of the first 20 terms

Find the sum of the first 20 terms of 3, 7, 11, 15, \ldots

d = 7 - 3 = 4

a_{20} = 3 + (20 - 1) \cdot 4

a_{20} = 3 + 19 \cdot 4

a_{20} = 3 + 76

a_{20} = 79

S_{20} = \frac{20(3 + 79)}{2}

S_{20} = \frac{20 \cdot 82}{2}

S_{20} = \frac{1640}{2}

S_{20} = 820

Answer: 820

Example 3: A sum in sigma notation

Evaluate \sum_{n=1}^{15} (2n + 5).

15 - 1 + 1 = 15

2(1) + 5 = 7

2(15) + 5 = 35

S_{15} = \frac{15(7 + 35)}{2}

S_{15} = \frac{15 \cdot 42}{2}

S_{15} = \frac{630}{2}

S_{15} = 315

Answer: 315. The terms go up by 2 each time, so this is an arithmetic series.

Example 4: Find the number of terms

How many terms of 5 + 8 + 11 + \cdots add up to 220?

Identify the parts: a_1 = 5, d = 3, S_n = 220

220 = \frac{n}{2}[2(5) + (n - 1) \cdot 3]

220 = \frac{n}{2}[10 + 3n - 3]

220 = \frac{n}{2}(3n + 7)

440 = n(3n + 7)

440 = 3n^2 + 7n

0 = 3n^2 + 7n - 440

0 = (3n + 40)(n - 11)

Solve each factor: n = -\frac{40}{3} or n = 11

The number of terms must be a positive whole number, so n = 11

Check: the 11th term is 5 + 10 \cdot 3 = 35, and \frac{11(5 + 35)}{2} = 220

Answer: 11 terms

Example 5: Word problem with yearly raises

A job pays $42,000 the first year, with a raise of $1,500 every year. What is the total pay for the first 10 years?

Identify the parts: a_1 = 42{,}000, d = 1500, n = 10

a_{10} = 42{,}000 + (10 - 1) \cdot 1500

a_{10} = 42{,}000 + 9 \cdot 1500

a_{10} = 42{,}000 + 13{,}500

a_{10} = 55{,}500

S_{10} = \frac{10(42{,}000 + 55{,}500)}{2}

S_{10} = \frac{10 \cdot 97{,}500}{2}

S_{10} = \frac{975{,}000}{2}

S_{10} = 487{,}500

Answer: The total pay for 10 years is $487,500.

Common Mistakes

Related Methods

Sum of the first n whole numbers

A handy special case is 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}. For example, 1 + 2 + \cdots + 8 = \frac{8 \cdot 9}{2} = 36.

Using the properties of sigma notation

You can also split a sum into simpler sums, as shown in sigma notation. For \sum_{n=1}^{15} (2n + 5), use 2 \cdot \frac{15 \cdot 16}{2} + 15 \cdot 5 = 240 + 75 = 315.

Average times count

The formula says the sum equals the number of terms times the average of the first and last terms. In Example 5, the average yearly pay is $48,750, and 10 \cdot 48{,}750 = 487{,}500.

Practice Problems

1. Find 1 + 2 + 3 + \cdots + 50.
2. Find the sum of the first 12 terms of 4, 9, 14, \ldots
3. Find the sum of the first 15 terms of 10, 7, 4, \ldots
4. Evaluate \sum_{k=1}^{25} (4k - 1).
5. Find the sum of the first 30 odd numbers.
6. Find the sum of all the multiples of 6 from 6 to 120.
7. Find 2 + 5 + 8 + \cdots + 62.
8. An arithmetic series has a_1 = 8 and d = 2. How many terms add up to 170?
9. An arithmetic series has 10 terms, a common difference of 3 and a sum of 245. Find the first term.
10. A theater has 24 seats in the first row, and each row has 3 more seats than the row in front of it. How many seats are in the first 15 rows?

S_{50} = \frac{50(1 + 50)}{2}

S_{50} = \frac{50 \cdot 51}{2}

S_{50} = \frac{2550}{2}

S_{50} = 1275

d = 9 - 4 = 5

a_{12} = 4 + 11 \cdot 5 = 4 + 55 = 59

S_{12} = \frac{12(4 + 59)}{2}

S_{12} = \frac{12 \cdot 63}{2}

S_{12} = \frac{756}{2}

S_{12} = 378

d = 7 - 10 = -3

a_{15} = 10 + 14 \cdot (-3) = 10 - 42 = -32

S_{15} = \frac{15(10 + (-32))}{2}

S_{15} = \frac{15 \cdot (-22)}{2}

S_{15} = \frac{-330}{2}

S_{15} = -165

4(1) - 1 = 3

4(25) - 1 = 99

S_{25} = \frac{25(3 + 99)}{2}

S_{25} = \frac{25 \cdot 102}{2}

S_{25} = \frac{2550}{2}

S_{25} = 1275

a_{30} = 1 + 29 \cdot 2 = 59

S_{30} = \frac{30(1 + 59)}{2}

S_{30} = \frac{30 \cdot 60}{2}

S_{30} = \frac{1800}{2}

S_{30} = 900

\frac{120}{6} = 20

S_{20} = \frac{20(6 + 120)}{2}

S_{20} = \frac{20 \cdot 126}{2}

S_{20} = \frac{2520}{2}

S_{20} = 1260

2 + (n - 1) \cdot 3 = 62

(n - 1) \cdot 3 = 60

n - 1 = 20

n = 21

S_{21} = \frac{21(2 + 62)}{2}

S_{21} = \frac{21 \cdot 64}{2}

S_{21} = \frac{1344}{2}

S_{21} = 672

170 = \frac{n}{2}[2(8) + (n - 1) \cdot 2]

170 = \frac{n}{2}[16 + 2n - 2]

170 = \frac{n}{2}(2n + 14)

170 = n(n + 7)

170 = n^2 + 7n

0 = n^2 + 7n - 170

0 = (n + 17)(n - 10)

245 = \frac{10}{2}[2a_1 + (10 - 1) \cdot 3]

245 = 5[2a_1 + 9 \cdot 3]

245 = 5(2a_1 + 27)

49 = 2a_1 + 27

22 = 2a_1

a_1 = 11

a_{15} = 24 + 14 \cdot 3 = 24 + 42 = 66

S_{15} = \frac{15(24 + 66)}{2}

S_{15} = \frac{15 \cdot 90}{2}

S_{15} = \frac{1350}{2}

S_{15} = 675

FAQ

What is the formula for an arithmetic series?

The sum of the first n terms is S_n = \frac{n(a_1 + a_n)}{2}. If you know the common difference but not the last term, use S_n = \frac{n}{2}[2a_1 + (n - 1)d].

What is the difference between an arithmetic sequence and an arithmetic series?

A sequence is a list of terms, like 3, 7, 11, 15. A series is their sum, like 3 + 7 + 11 + 15 = 36.

How do you find the sum of the numbers from 1 to 100?

Pair the first and last numbers: each pair adds to 101, and there are 50 pairs. So the sum is 50 \cdot 101 = 5050.

Can an arithmetic series have a negative sum?

Yes. If the common difference is negative, the terms eventually become negative. For example, the first 15 terms of 10, 7, 4, \ldots add up to −165.

Summary

• An arithmetic series adds the terms of an arithmetic sequence.
• S_n = \frac{n(a_1 + a_n)}{2}: the number of terms times the average of the first and last terms.
• S_n = \frac{n}{2}[2a_1 + (n - 1)d] when you know d but not the last term.
• Find the last term with a_n = a_1 + (n - 1)d.
• The number of terms is always a positive whole number.

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