UNIT 12.2/12.4 ARITHMETIC UNIT 12.2/12.4 ARITHMETIC SEQUENCES AND SERIESSEQUENCES AND SERIES

Warm UpFind the 5th term of each sequence.

1. an = n + 6 2. an = 4 – n

3. an = 3n + 4Write a possible explicit rule for the nth term of each sequence.

4. 4, 5, 6, 7, 8,… 5. –3, –1, 1, 3, 5, …

6.

11 –1

19

an = n + 3 an = 2n – 5

Find the indicated terms of an arithmetic sequence.

Find the sums of arithmetic series.

Objectives

arithmetic sequencearithmetic series

Vocabulary

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. 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.

Check It Out! Example 1a

Determine whether the sequence could be arithmetic. If so, find the common difference and the next term.

1.9, 1.2, 0.5, –0.2, –0.9, ...

The sequence could be arithmetic with a common difference of –0.7.

1.9, 1.2, 0.5, –0.2, –0.9

–0.7 –0.7 –0.7 –0.7 Differences

The next term would be –0.9 – 0.7 = –1.6.

The sequence is not arithmetic because the first differences are not common.

Check It Out! Example 1b

Determine whether the sequence could be arithmetic. If so, find the common difference and the next term.

Differences

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. 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.

Check It Out! Example 2a

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.

a11= –3 + (11 – 1)(–2) Substitute –3 for a1, 11 for n, and –2 for d.

= –23

The 11th term is –23.

Check It Out! Example 2a Continued

n 1 2 3 4 5 6 7 8 9 10 11

an –11 –13 –15 –17 –19 –21 –23 –9 –7 –5 –3

Check Continue the sequence.

Find the 11th term of the arithmetic sequence.

Check It Out! Example 2b

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.

a11= 9.2 + (11 – 1)(–0.05) Substitute 9.2 for a1, 11 for n, and –0.05 for d.

= 8.7

The 11th term is 8.7.

Check It Out! Example 2b Continued

Check Continue the sequence. n 1 2 3 4 5 6 7 8 9 10 11

an 9.2 9.15 9.1 9.05 9 8.95 8.9 8.85 8.8 8.75 8.7

Check It Out! Example 3

Find the missing terms in the arithmetic sequence

2, , , , 0.

an = a1 + (n – 1)d

0 = 2 + (5 – 1)d

–2 = 4d

Step 1 Find the common difference.

General rule.

Substitute 0 for an, 2 for a1, and 5 for n.

Solve for d.

= 1

Check It Out! Example 3 Continued

The missing terms are

Step 2 Find the missing terms using d= and a1= 2.

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

Check It Out! Example 4a

Find the 11th term of the arithmetic sequence.

a2 = –133 and a3 = –121

an = a1 + (n – 1)d

a3 = a2 + (3 – 2)d

–121 = –133 + d

d = 12

Step 1 Find the common difference.

Let an = a3 and a1 = a2. Replace 1 with 2.

Simplify.

Substitute –121 for a3 and –133 for a2.

a3 = a2 + d

Check It Out! Example 4a Continued

an = a1 + (n – 1)d

–133 = a1 + (2 – 1)(12)

–133 = a1 + 12

–145 = a1

Step 2 Find a1.

General rule

Substitute –133 for an, 2 for n, and 12 for d.

Simplify.

an = a1 + (n – 1)d

a11 = –145 + (n – 1)(12)

= –25

The 11th term is –25.

Step 3 Write a rule for the sequence, and evaluate to find a11.

General rule.

Substitute –145 for a1 and 12 for d.

Evaluate for n = 11.

Check It Out! Example 4a Continued

a11 = –145 + (11 – 1)(12)

Check It Out! Example 4b

Find the 11th term of each arithmetic sequence.

a3 = 20.5 and a8 = 13

an = a1 + (n – 1)d

a8 = a3 + (8 – 3)d

Step 1 Find the common difference.

a8 = a3 + 5d

13 = 20.5 + 5d

–7.5 = 5d

–1.5 = d

General ruleLet an = a8 and a1 = a3.

Replace 1 with 3.

Simplify.

Substitute 13 for a8 and 20.5 for a3.

Simplify.

Check It Out! Example 4b Continued

an = a1 + (n – 1)d

20.5 = a1 + (3 – 1)(–1.5)

20.5 = a1 – 3

23.5 = a1

Step 2 Find a1.

General rule

Substitute 20.5 for an, 3 for n, and –1.5 for d.

Simplify.

Check It Out! Example 4b Continued

an = a1 + (n – 1)d

a11 = 23.5 + (n – 1)(–1.5)

Step 3 Write a rule for the sequence, and evaluate to find a11.

a11 = 8.5

The 11th term is 8.5.

General rule

Substitute 23.5 for a1 and –1.5 for d.

Evaluate for n = 11.a11 = 23.5 + (11 – 1)(–1.5)

In Lesson 12-2 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.

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!

Check It Out! Example 5a

Find the indicated sum for the arithmetic series.

S16 for 12 + 7 + 2 +(–3)+ …

d = 7 – 12 = –5

a16 = 12 + (16 – 1)(–5)

= –63

Find the common difference.

Find the 16th term.

Check It Out! Example 5a Continued

= 16(–25.5)

= –408

Substitute.

Sum formula.

Simplify.

Find S16.

Check It Out! Example 5b

Find the indicated sum for the arithmetic series.

a1 = 50 – 20(1) = 30

a15 = 50 – 20(15) = –250

Find 1st and 15th terms.

= 15(–110)

= –1650

Check It Out! Example 5b Continued

Find S15.

Substitute.

Sum formula.

Simplify.

Check It Out! Example 6a What if...? The number of seats in the first row of a theater has 14 seats. Suppose that each row after the first had 2 additional seats.

How many seats would be in the 14th row?Write a general rule using a1 = 14 and d = 2.

an = a1 + (n – 1)d

a14 = 11 + (14 – 1)(2)

= 11 + 26= 37

Explicit rule for nth term

Substitute.

Simplify.

There are 37 seats in the 14th row.

Check It Out! Example 6b

How many seats in total are in the first 14 rows?

Find S14 using the formula for finding the sum of the first n terms.

There are 336 total seats in rows 1 through 14.

Formula for first n terms

Substitute.

Simplify.

All rights belong to their respective owners.Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "fair use" for purposes such as criticism, comment, news reporting, TEACHING, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, EDUCATIONAL or personal use tips the balance in favor of fair use.