9-3 9-3 Arithmetic ArithmeticSequences Sequencesand andSeries Series Warm Up Lesson Presentation Lesson Quiz HoltMcDougal Algebra 2Algebra 2 Holt 9-3 Arithmetic Sequences and Series Warm Up Find the 5th term of each sequence. 1. an = n + 6 11 2. an = 4 – n –1 3. an = 3n + 4 19 Write a possible explicit rule for the nth term of each sequence. an = n + 3 4. 4, 5, 6, 7, 8,… Holt McDougal Algebra 2 an = 2n – 5 5. –3, –1, 1, 3, 5, … 9-3 Arithmetic Sequences and Series Objectives Find the indicated terms of an arithmetic sequence. Find the sums of arithmetic series. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Vocabulary arithmetic sequence arithmetic series Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Essential Question • How do you find the sum of an arithmetic series? Holt McDougal Algebra 2 9-3 Arithmetic Sequences and 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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, Differences –4, 6 2, 6 8, 6 14 6 The sequence could be arithmetic with a common difference of 6. The next term is 14 + 6 = 20. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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, Differences –5, –3 –11, –6 –20, –32 –9 –12 The sequence is not arithmetic because the first differences are not common. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Example 2 Continued Step 2 Evaluate by using the formula. an = a1 + (n – 1)d a12 General rule. = 20 + (12 – 1)(–6) Substitute 20 for a1, 12 for n, and –6 for d. = –46 The 12th term is –46. Check Continue the sequence. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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) = –23 The 11th term is –23. Holt McDougal Algebra 2 Substitute –3 for a1, 11 for n, and –2 for d. 9-3 Arithmetic Sequences and Series Check It Out! Example 2a Continued Check Continue the sequence. n 1 2 3 4 5 6 7 8 9 10 11 an –3 –5 –7 –9 –11 –13 –15 –17 –19 –21 –23 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. –7 = 17 + (5 – 1)(d) Substitute –7 for an, 17 for a1, and 5 for n. –6 = d Solve for d. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Example 3 Continued Step 2 Find the missing terms using d= –6 and a1 = 17. a2 = 17 + (2 – 1)(–6) = 11 a3 = 17 +(3 – 1)(–6) =5 a4 = 17 + (4 – 1)(–6) = –1 Holt McDougal Algebra 2 The missing terms are 11, 5, and –1. 9-3 Arithmetic Sequences and Series Check It Out! Example 3 Find the missing terms in the arithmetic sequence 2, , , , 0. Step 1 Find the common difference. an = a1 + (n – 1)d 0 = 2 + (5 – 1)d –2 = 4d General rule. Substitute 0 for an, 2 for a1, and 5 for n. Solve for d. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 3 Continued Step 2 Find the missing terms using d= and a1= 2. The missing terms are =1 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Because arithmetic sequences have a common difference, you can use any two terms to find the difference. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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 a14 = a8 + (14 – 8)d Let an = a14 and a1 = a8. Replace 1 with 8. a14 = a8 + 6d Simplify. 157 = 85 + 6d 72 = 6d 12 = d Holt McDougal Algebra 2 Substitute 157 for a14 and 85 for a8. 9-3 Arithmetic Sequences and Series Example 4 Continued Step 2 Find a1. an = a1 + (n – 1)d General rule 85 = a1 + (8 - 1)(12) Substitute 85 for a8, 8 for n, and 12 for d. 85 = a1 + 84 Simplify. 1 = a1 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. a5 = 1 + (5 – 1)(12) Evaluate for n = 5. = 49 The 5th term is 49. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 4a Find the 11th term of the arithmetic sequence. a2 = 25 and a4 = 39 Step 1 Find the common difference. an = a1 + (n – 1)d a4 = a2 + (4 – 2)d Let an = a4 and a1 = a2. Replace 1 with 2. a4 = a2 + 2d 39 = 25 + 2d d=7 Holt McDougal Algebra 2 Simplify. Substitute 39 for a4 and 25 for a2. 9-3 Arithmetic Sequences and Series Check It Out! Example 4a Continued Step 2 Find a1. an = a1 + (n – 1)d General rule 25 = a1 + (2 – 1)(7) Substitute 25 for an, 2 for n, and 7 for d. 25 = a1 + 7 Simplify. 18 = a1 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 4a Continued Step 3 Write a rule for the sequence, and evaluate to find a11. an = a1 + (n – 1)d General rule. a11 = 18 + (n – 1)(7) Substitute 18 for a1 and 7 for d. a11 = 18 + (11 – 1)(7) Evaluate for n = 11. = 88 The 11th term is 88. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series In Lesson 9-1 you wrote and evaluated sequences. An arithmetic series is the indicated sum of the terms of an arithmetic sequence. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Find Sn for an Arithmetic Series. a1 = 1, an = 19, n = 18 a1 a18 Sn n 2 Sum Formula Substitute 18 for n, 1 for a1 , and 19 for a18 S18 = 18(20/2) S18 = 18(10) S18 = 180 Holt McDougal Algebra 2 Simplify. 9-3 Arithmetic Sequences and Series 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 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Example 5A Continued a1 a18 Sn n 2 Sum formula Substitute. = 18(-80.5) = –1449 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Remember! 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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 a15 Sn n 2 a1 = 5 + 2(1) = 7 a15 = 5 + 2(15) = 35 = 15(21) = 315 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 5a Continued Find S16. a1 a16 Sn n 2 Sum formula. Substitute. = 16(–25.5) = –408 Holt McDougal Algebra 2 Simplify. 9-3 Arithmetic Sequences and Series Check It Out! Example 5b Find the indicated sum for the arithmetic series. Find 1st and 15th terms. a1 = 50 – 20(1) = 30 a15 = 50 – 20(15) = –250 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 5b Continued Find S15. a1 a15 Sn n 2 Sum formula. Substitute. = 15(–110) = –1650 Holt McDougal Algebra 2 Simplify. 9-3 Arithmetic Sequences and Series 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 a20 = 15 + (20 – 1)(2) Explicit rule for nth term Substitute. = 15 + 38 Simplify. = 53 There are 53 seats in the 20th row. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 6a What if...? The number of seats in the first row of a theater has 11 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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. Formula for first n terms Substitute. Simplify. There are 336 total seats in rows 1 through 14. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Lesson Quiz: Part I 1. Determine whether the sequence could be arithmetic. If so, find the first difference and the next term. –1, –4, –7, –10, –13, … yes; –3,–16 2. Find the 10th term of the arithmetic sequence –2, –5, –8, –11, –14, … –29 3. Find the missing terms in the arithmetic sequence 15, , , , 23. 17, 19, 21 4. Find the sum of the arithmetic sequence with a1 = 11 and a12 = -33. -132 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Lesson Quiz: Part II 5. Find the indicated sum for –132 6. The side section of an auditorium has 12 seats in the first row and 3 additional seats in each subsequent row. How many seats are in the 10th row? How many seats in total are in the first 10 rows? 39; 255 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Essential Question • How do you find the sum of an arithmetic series? Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series • Teacher: You have the terms of the sequence all mixed up. • Student: If you wanted them in order, why didn’t you just say so? Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series 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, ... 1.9, Differences 1.2, –0.7 0.5, –0.7 –0.2, –0.7 –0.9 –0.7 The sequence could be arithmetic with a common difference of –0.7. The next term would be –0.9 – 0.7 = –1.6. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 1b Determine whether the sequence could be arithmetic. If so, find the common difference and the next term. Differences The sequence is not arithmetic because the first differences are not common. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 2b 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. 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. Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 2b Continued Check Continue the sequence. n 1 an 9.2 2 3 4 9.15 9.1 9.05 Holt McDougal Algebra 2 5 6 7 8 9 10 11 9 8.95 8.9 8.85 8.8 8.75 8.7 9-3 Arithmetic Sequences and Series Check It Out! Example 4b 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 a8 = a3 + (8 – 3)d General rule Let an = a8 and a1 = a3. Replace 1 with 3. a8 = a3 + 5d Simplify. 13 = 20.5 + 5d Substitute 13 for a8 and 20.5 for a3. –7.5 = 5d –1.5 = d Holt McDougal Algebra 2 Simplify. 9-3 Arithmetic Sequences and Series Check It Out! Example 4b Continued Step 2 Find a1. an = a1 + (n – 1)d General rule 20.5 = a1 + (3 – 1)(–1.5) Substitute 20.5 for an, 3 for n, and –1.5 for d. 20.5 = a1 – 3 Simplify. 23.5 = a1 Holt McDougal Algebra 2 9-3 Arithmetic Sequences and Series Check It Out! Example 4b Continued Step 3 Write a rule for the sequence, and evaluate to find a11. an = a1 + (n – 1)d a11 = 23.5 + (n – 1)(–1.5) a11 = 23.5 + (11 – 1)(–1.5) a11 = 8.5 The 11th term is 8.5. Holt McDougal Algebra 2 General rule Substitute 23.5 for a1 and –1.5 for d. Evaluate for n = 11. 9-3 Arithmetic Sequences and Series Example 5B Continued Check Use a graphing calculator. Holt McDougal Algebra 2
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