Introduction to Arithmetic Sequences
A term in an arithmetic sequence is computed by adding a fixed number to the previous term. For example, 3, 7, 11, 15, 19, ... is an arithmetic sequence because we can add 4 to any term to find the following term. We can define the n th term recursively as a n = a n −1 + d or, in more general terms, a n = a 1 + ( n − 1) d . In the sequence above, a 1 = 3 and d = 4.
Examples
Find the first four terms and the 100th term.

When asked whether or not a sequence is arithmetic, we will find the difference between consecutive terms. If the difference is the same, the sequence is arithmetic.
Examples
Determine if the sequence is arithmetic. If it is, find the common difference.
- −8, −1, 6, 13, 20, ...
-
20 − 13 = 7, 13 − 6 = 7, 6 − (−1) = 7, −1 − (−8) = 7
The sequence is arithmetic. The common difference is 7.
- 29, 17, 5, −7, −19, ...
-
−19 − (−7) = −12, −7 − 5 = −12, 5 − 17 = −12, 17 − 29 = −12
The sequence is arithmetic, and the common difference is −12.
Because the differences are not the same, the sequence is not arithmetic.
We can find any term in an arithmetic sequence if we know either one term and the common difference or two terms. We need to use the formula a n = a 1 + ( n − 1) d and, if necessary, a little algebra. For example, if we are told the common difference is 6 and the tenth term is 141, then we can put a n = 141, n = 10, and d = 6 in the formula to find a 1 .
141 = a 1 + (10 − 1)6
87 = a 1
The n th term is a n = 87 + ( n − 1)6.
Examples
Find the n th term for the arithmetic sequence.
- The common difference is 2/3 and the seventh term is −10.
-
Using
, n = 7, and a n = −10, the formula a n = a 1 + ( n − 1) d becomes
.

The n th term is
.
- The twelfth term is 8, and the twentieth term is 32.
-
The information gives us a system of two equations with two variables. In this example and the rest of the problems in this section, we will add −1 times the first equation to the second. Substitution and matrices would work, too. The equations are 8 = a 1 + (12 − 1) d and 32 = a 1 + (20 − 1) d .

The n th term is a n = −25 + ( n −1)3.
- The eighth term is 4, and the twentieth term is −38.
-
The information in these two terms gives us the system of equations 4 = a 1 + (8 − 1) d and −38 = a 1 + (20 − 1) d .

The n th term is
.
We can add the first n terms of an arithmetic sequence using one of the following two formulas.

We will use the first formula if we know all of a 1 , a n , and n , and the second if we do not know a n .
Examples
- Find the sum.

a 1 = 2, a 6 = 5, and n = 6 (because there are six terms)

- Find the sum of the first 20 terms of the sequence −5, − 1, 3, 7, 11, .... a 1 = −5, d = 4, and n = 20.

- 6 + (−2) + (−10) + (−18) + ... + (−58)
-
We know a 1 = 6, d = −8 and a n = −58 but not n . We can find n by solving −58 = 6 + ( n − 1)(−8).
−58 = 6 + ( n − 1)(−8)
−64 = −8( n − 1)
8 = n − 1
9 = n

- Find the sum of the first thirty terms of the arithmetic sequence whose fifth term is 19 and whose tenth term is 31.5.
-
In order to use the second sum formula, we need to find a 1 and d . If we were to use the first formula, we would have to find a 30, which is a little more work. Because a 5 = 19 and a 10 = 31.5, we have the system of equations 19 = a 1 + (5 − 1) d and 31.5 = a 1 + (10 − 1) d .

a 1 = 9

-
1
- 2
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. The n th term is
.
, and n = 6.
becomes S 6 = 

. The information in the problem gives the system 8 = a 1 + (4 − 1) d and 56 = a 1 + (10 − 1) d.

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