1. Lesson 12–3 Objectives
Be able to find the terms of an ARITHMETIC sequence
Be able to find the sums of arithmetic series

2. Arithmetic sequence – when the terms differ by the same number
This is called the common difference, “d”

3. Example 1: Identifying Arithmetic Sequences
Determine whether the sequence could be arithmetic. If so, find the common first
difference and the next term.
–10, –4, 2, 8, 14, …
–10, –4, , 8, 14
Differences
The sequence could be arithmetic with a common difference of 6.
The next term is = 20.

4. Example 1: Identifying Arithmetic Sequences
Determine whether the sequence could be arithmetic. If so, find the common first
difference and the next term.
–2, –5, –11, –20, –32, …
–2, –5, –11, –20, –32
Differences – – – –12
The sequence is not arithmetic because the first differences are not common.

5. Example 1: Identifying Arithmetic Sequences
Determine whether the sequence could be arithmetic. If so, find the common first
difference and the next term.
1.9, 1.2, 0.5, –0.2, –0.9, ...
1.9, , , –0.2, –0.9
–0.7
Differences
The sequence could be arithmetic with a common difference of –0.7.
The next term would be –0.9 – 0.7 = –1.6.

6. Formulas for arithmetic sequences:
Recursive formula: an = an–1 + d
Explicit formula: an = a1 + (n – 1)d

7. Example 2: Finding the nth Term Given an Arithmetic Sequence
Find the 12th term of the arithmetic sequence 20, 14, 8, 2, 4, ....
Step 1 Find the common difference: d = 14 – 20 = –6.

8.  Step 2 Evaluate by using the formula. an = a1 + (n – 1)d
General rule.
Substitute 20 for a1, 12 for n, and –6 for d.
a12 = 20 + (12 – 1)(–6)
= –46
The 12th term is –46.
Check Continue the sequence. 

9. Example 2: Finding the nth Term Given an Arithmetic Sequence
Find the 11th term of the arithmetic sequence.
–3, –5, –7, –9, …
Step 1 Find the common difference: d = –5 – (–3)= –2.
Step 2 Evaluate by using the formula.
an = a1 + (n – 1)d
General rule.
Substitute –3 for a1, 11 for n, and –2 for d.
a11= –3 + (11 – 1)(–2)
= –23
The 11th term is –23.

10. Example 2: Finding the nth Term Given an Arithmetic Sequence
Find the 11th term of the arithmetic sequence.
9.2, 9.15, 9.1, 9.05, …
Step 1 Find the common difference:
d = 9.15 – 9.2 = –0.05.
Step 2 Evaluate by using the formula.
an = a1 + (n – 1)d
General rule.
Substitute for a1, 11 for n, and –0.05 for d.
a11= (11 – 1)(–0.05)
= 8.7
The 11th term is 8.7.

11. Example 3: Finding Missing Terms
Find the missing terms in the arithmetic sequence
17, , , , –7.
Step 1 Find the common difference.
an = a1 + (n – 1)d
General rule.
Substitute –7 for an,
17 for a1, and 5 for n.
–7 = 17 + (5 – 1)(d)
–6 = d
Solve for d.

12. Step 2 Find the missing terms using d= –6 and a1 = 17.
= 11
The missing terms are 11, 5, and –1.
a3 = 17 +(3 – 1)(–6)
= 5
a4 = 17 + (4 – 1)(–6)
= –1

13. Example 3: Finding Missing Terms
Find the missing terms in the arithmetic sequence
2, , , , 0.
Step 1 Find the common difference.
an = a1 + (n – 1)d
General rule.
0 = 2 + (5 – 1)d
Substitute 0 for an,
2 for a1, and 5 for n.
–2 = 4d
Solve for d.

14. Step 2 Find the missing terms using d= and a1= 2.
The missing terms are
= 1

15. You can use the explicit formula even if you don’t know the first term…

16. Example 4: Finding the nth Term Given Two Terms
Find the 5th term of the arithmetic sequence with a8 = 85 and a14 = 157.
Step 1 Find the common difference.
Let an = a14 and a1 = a8. Replace 1 with 8.
a14 = a8 + (14 – 8)d
a14 = a8 + 6d
Simplify.
Substitute 157 for a14 and 85 for a8.
157 = d
72 = 6d
12 = d

17. Step 2 Write a rule for the sequence, and evaluate to find a5.
Repeat the process again, this time letting “a5” be the unknown
a14 = a5 + (14 – 5)d
157 = a5 + (14 – 5)12
157 = a
49 = a5
The 5th term is 49.

18. Example 4: Finding the nth Term Given Two Terms
Find the 11th term of the arithmetic sequence.
a2 = –133 and a3 = –121
Step 1 Find the common difference.
an = a1 + (n – 1)d
a3 = a2 + (3 – 2)d
Let an = a3 and a1 = a2. Replace 1 with 2.
a3 = a2 + d
Simplify.
–121 = –133 + d
Substitute –121 for a3 and –133 for a2.
d = 12

19. Step 2 Write a rule for the sequence, and evaluate to find a11.
Repeat the process again, this time letting “a11” be the unknown
a3 = a11 + (3 – 11)d
–121 = a11 + (3 – 11)12
–121 = a5 – 96
–25 = a11
The 11th term is -25.

20. Example 4: Finding the nth Term Given Two Terms
Find the 11th term of each arithmetic sequence.
a3 = 20.5 and a8 = 13
Step 1 Find the common difference.
an = a1 + (n – 1)d
General rule
Let an = a8 and a1 = a3.
Replace 1 with 3.
a8 = a3 + (8 – 3)d
a8 = a3 + 5d
Simplify.
13 = d
Substitute 13 for a8 and 20.5 for a3.
–7.5 = 5d
Simplify.
–1.5 = d

21. Step 2 Write a rule for the sequence, and evaluate to find a11.
Repeat the process again, this time letting “a11” be the unknown
a8 = a11 + (8 – 11)d
13 = a11 + (8 – 11)(–1.5)
13 = a
8.5 = a11
The 11th term is 8.5.

22. Arithmetic series – the sum of specified terms in an arithmetic sequence

23. Example 5: Finding the Sum of an Arithmetic Series
Find the indicated sum for the arithmetic series.
S18 for (–9) + (–20)
Find the common difference.
d = 2 – 13 = –11
Find the 18th term.
a18 = 13 + (18 – 1)(–11)
= –174

24.  Sum formula Substitute. = 18(-80.5) = –1449
Check Use a graphing calculator. 

25. Example 5: Finding the Sum of an Arithmetic Series
Find the indicated sum for the arithmetic series.
Find S15.
Find 1st and 15th terms.
a1 = 5 + 2(1) = 7
a15 = 5 + 2(15) = 35
= 15(21) = 315

26. Check Use a graphing calculator.

27. Example 5: Finding the Sum of an Arithmetic Series
Find the indicated sum for the arithmetic series.
S16 for (–3)+ …
Find the common difference.
d = 7 – 12 = –5
Find the 16th term.
a16 = 12 + (16 – 1)(–5)
= –63

28. Find S16.
Sum formula.
Substitute.
= 16(–25.5)
Simplify.
= –408

29. Example 5: Finding the Sum of an Arithmetic Series
Find the indicated sum for the arithmetic series.
Find 1st and 15th terms.
a1 = 50 – 20(1) = 30
a15 = 50 – 20(15) = –250

30. Find S15.
Sum formula.
Substitute.
= 15(–110)
Simplify.
= –1650

31. Example 6: Theater Application
The center section of a concert hall has 15 seats in the first row and 2 additional seats in each subsequent row.
How many seats are in the 20th row?
Write a general rule using a1 = 15 and d = 2.
an = a1 + (n – 1)d
Explicit rule for nth term
a20 = 15 + (20 – 1)(2)
Substitute. =
Simplify.
= 53
There are 53 seats in the 20th row.

32. How many seats in total are in the first 20 rows?
Find S20 using the formula for finding the sum of the first n terms.
Formula for first n terms
Substitute.
Simplify.
There are 680 seats in rows 1 through 20.

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