Objectives Find the indicated terms of an arithmetic sequence. Find the sums of arithmetic series.

The cost of mailing a letter in 2005 gives the sequence 0.37, 0.60, 0.83, 1.06, …. This sequence is called an arithmetic sequence because its successive terms differ by the same number d (d ≠ 0), called the common difference. For the mail costs, d is 0.23, as shown.

Recall that linear functions have a constant first difference Recall that linear functions have a constant first difference. Notice also that when you graph the ordered pairs (n, an) of an arithmetic sequence, the points lie on a straight line. Thus, you can think of an arithmetic sequence as a linear function with sequential natural numbers as the domain. Common difference (d)  

Example 1A: 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, 2, 8, 14 Differences 6 6 6 6 The sequence could be arithmetic with a common difference of 6. The next term is 14 + 6 = 20.

Example 1B: 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 –3 –6 –9 –12 The sequence is not arithmetic because the first differences are not common.

Each term in an arithmetic sequence is the sum of the previous term and the common difference. This gives the recursive rule an = an – 1 + d. You also can develop an explicit rule for an arithmetic sequence.

Notice the pattern in the table Notice the pattern in the table. Each term is the sum of the first term and a multiple of the common difference. This pattern can be generalized into a rule for all arithmetic sequences.

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

 Example 2 Continued 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. 

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.

Example 3 Continued Step 2 Find the missing terms using d= –6 and a1 = 17. a2 = 17 + (2 – 1)(–6) = 11 The missing terms are 11, 5, and –1. a3 = 17 +(3 – 1)(–6) = 5 a4 = 17 + (4 – 1)(–6) = –1

Because arithmetic sequences have a common difference, you can use any two terms to find the difference.

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. an = a1 + (n – 1)d 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 = 85 + 6d 72 = 6d 12 = d

Example 4 Continued Step 2 Find a1, given: d=12 and n=8 an = a1 + (n – 1)d General rule Substitute 85 for a8, 8 for n, and 12 for d. 85 = a1 + (8 - 1)(12) 85 = a1 + 84 Simplify. 1 = a1

Example 4 Continued Step 3 Write a rule for the sequence, and evaluate to find a5. an = a1 + (n – 1)d General rule. an = 1 + (n – 1)(12) Substitute 1 for a1 and 12 for d. an = 1 + 12n – 12 Distribute 12 an = 12n – 11 Explicit rule.

In Lesson 9-2 you wrote and evaluated series In Lesson 9-2 you wrote and evaluated series. An arithmetic series is the indicated sum of the terms of an arithmetic sequence. You can derive a general formula for the sum of an arithmetic series by writing the series in forward and reverse order and adding the results.

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

Example 5A Continued Sum formula Substitute. = 18(-80.5) = –1449 These sums are actually partial sums. You cannot find the complete sum of an infinite arithmetic series because the term values increase or decrease indefinitely. Remember!

Example 5B: 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

Check It Out! Example 5a Find the indicated sum for the arithmetic series. S16 for 12 + 7 + 2 +(–3)+ … Find the common difference. d = 7 – 12 = –5 Find the 16th term. a16 = 12 + (16 – 1)(–5) = –63

Check It Out! Example 5a Continued Find S16. Sum formula. Substitute. = 16(–25.5) Simplify. = –408

Example 6A: 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. = 15 + 38 Simplify. = 53 There are 53 seats in the 20th row.

Example 6B: Theater Application 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.