1. Arithmetic Sequences and Series
LESSON 10–2
Arithmetic Sequences and Series

2. Five-Minute Check (over Lesson 10-1) TEKS Then/Now New Vocabulary
Example 1: Arithmetic Sequences
Key Concept: The nth Term of an Arithmetic Sequence
Example 2: Explicit and Recursive Formulas
Example 3: nth Terms
Example 4: Arithmetic Means
Example 5: Second Differences
Key Concept: Sum of a Finite Arithmetic Series
Example 6: Sums of Arithmetic Series
Example 7: Real-World Example: Sum of an Arithmetic Series
Example 8: Real-World Example: Sum of an Arithmetic Series
Lesson Menu

3. Find the next four terms of the sequence 2, 5, 8, 11, … .
A. 13, 15, 17, 19
B. 14, 17, 20, 23
C. 15, 19, 23, 27
D. 15, 20, 26, 33
5–Minute Check 1

4. Find the next four terms of the sequence 3, −9, 27, −81, … .
A. 162, –324, 648, –1296
B. 243, –729, 2187, –6561
C. 243, 729, 2187, 6561
D. 324, –1296, 5184, –20,736
5–Minute Check 2

5. Find the third term of the sequence a1 = 10, an = –2an–1 + 4.
B. a3 = –16
C. a3 = 36
D. a3 = 76
5–Minute Check 3

6. Find the sixth term of the sequence an = .
A. a6 =
B. a6 =
C. a6 =
D. a6 = 3
5–Minute Check 4

7. Find
A. 8
B. 23
C. 50
D. 90
5–Minute Check 5

8. Which of the following sequences is divergent?
A. 3, 11, 15, 17, …
B. 9, 7, 5, 3, …
C. 20, −5, 1.25, –0.3125, …
D. 32, 16, 8, 4, …
5–Minute Check 6

9. Targeted TEKS
P.5(C) Calculate the n th term and the n th partial
sum of an arithmetic series in mathematical and real-world problems.
P.5(D) Represent arithmetic series and geometric series using sigma notation.
Also addresses P.5(A) and P.5(B).
Mathematical Processes
P.1(B), P.1(G)

10. You found terms of sequences and sums of series. (Lesson 10-1)
Find nth terms and arithmetic means of arithmetic sequences.
Find sums of n terms of arithmetic series.
Then/Now

11. arithmetic sequence common difference arithmetic means
first difference
second difference
arithmetic series
Vocabulary

12. First, find the common difference.
Arithmetic Sequences
Determine the common difference and the next four terms of the arithmetic sequence –53, –36, –19, … .
First, find the common difference.
a2  a1 = –36 – (–53) or 17 Find the difference between pairs of consecutive terms to determine the common difference.
a3  a2 = –19 – (–36) or 17
Example 1

13. The next four terms are –2, 15, 32, and 49.
Arithmetic Sequences
The common difference is 17. Add 17 to the third term to find the fourth term, and so on.
a4 = – = –2
a5 = – = 15
a6 = = 32
a7 = = 49
The next four terms are –2, 15, 32, and 49.
Answer: 17; –2, 15, 32, 49
Example 1

14. Determine the common difference and find the next four terms of the arithmetic sequence 74, 68, 62, 56, … .
A. –6; 50, 44, 38, 32
B. –5; 51, 46, 41, 36
C. –4; 52, 48, 44, 40
D. 6; 62, 68, 74, 80
Example 1

15. Key Concept 2

16. First, find the common difference.
Explicit and Recursive Formulas
Find both an explicit formula and a recursive formula for the nth term of the arithmetic sequence 14, 3, –8, … .
First, find the common difference.
a2  a1 = 3  14 or 11 Find the difference between pairs of consecutive terms to determine the common difference.
a3  a2 = 8  3 or 11
Example 2

17. an = a1 + (n 1)d nth term of an arithmetic sequence
Explicit and Recursive Formulas
For an explicit formula, substitute a1 = 14 and d = –11 in the formula for the nth term of an arithmetic sequence.
an = a1 + (n 1)d nth term of an arithmetic sequence
= 14 + (n  1)(11) a1 = and d = –11
= 14 + (11)(n  1) or –11n + 25 Simplify.
Example 2

18. Explicit and Recursive Formulas
For a recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous term an  1 and d.
a1 = 14
an = an  1 + (11) or an = an  1 – 11
An explicit formula for the sequence is an = 14 + (11)(n  1) or an = –11n +25, and a recursive formula is a1 = 14, an = an  1 – 11.
Answer: an = 14 + (–11)(n – 1) or an = –11n + 25; a1 = 14, an = an – 1 – 11
Example 2

19. Find both an explicit formula and a recursive formula for the nth term of the arithmetic sequence 15, 33, 51, … .
A. an = 15n; a1 = 15, an = an –
B. an = 15n; a1 = 18, an = an –
C. an = 18n – 3; a1 = 15, an = an –
D. an = 18n; a1 = 15, an = an –
Example 2

20. A. Find the 41st term of the arithmetic sequence 11, 4, –3, –10, ….
nth Terms
A. Find the 41st term of the arithmetic sequence 11, 4, –3, –10, ….
First, find the common difference.
a2 – a1 = 4 – 11 or –7 Find the difference between pairs of consecutive terms to determine the common difference.
a3 – a2 = –3 – 4 or –7
Example 3

21. an = a1 + (n – 1)d nth term of an arithmetic sequence
nth Terms
Use the formula for the nth term of an arithmetic sequence to find a41.
an = a1 + (n – 1)d nth term of an arithmetic sequence
a41 = 11 + (41 – 1)(–7) n = 41, a1 = 11, and d = –7
a41 = –269 Simplify.
The 41st term of the sequence is –269.
Answer: –269
Example 3

22. Use the formula for the nth term of an arithmetic sequence.
nth Terms
B. Find the first term of the arithmetic sequence for which a44 = 229 and d = 8.
Use the formula for the nth term of an arithmetic sequence.
an = a1 + (n  1)d nth term of an arithmetic sequence
229 = a1 + (44  1)8 n = 44, an = 229, and d = 8
a1 = –115 Simplify.
The first term of the sequence is –115.
Answer: –115
Example 3

23. Find the 25th term of the arithmetic sequence 49, 40, 31, ….
B. –167
C. –158
D. 265
Example 3

24. an = a1 + (n  1)d nth term of an arithmetic sequence
Arithmetic Means
Write an arithmetic sequence that has five arithmetic means between 6.2 and –1.6.
The sequence will resemble 6.2, _____ , _____ , _____ , _____ , _____ , –1.6. Note that –1.6 is the 7th term of the sequence or a7. ?
First, find the common difference using a7 = –1.6, a1 = 6.2, and n = 7.
an = a1 + (n  1)d nth term of an arithmetic sequence
–1.6 = (7 – 1)d an = –1.6, a1 = 6.2, and n = 7
Example 4

25. Then determine the arithmetic means by using d = –1.3.
–1.6 = d Simplify.
d = –1.3 Solve for d.
Then determine the arithmetic means by using d = –1.3.
a2 = (–1.3) or 4.9
a3 = (–1.3) or 3.6
a4 = (–1.3) or 2.3
a5 = (–1.3) or 1
a6 = 1 + (–1.3) or –0.3
The sequence is 6.2, 4.9, 3.6, 2.3, 1, –0.3, –1.6.
Answer: 6.2, 4.9, 3.6, 2.3, 1, –0.3, –1.6
Example 4

26. Write an arithmetic sequence that has five arithmetic means between 16
Write an arithmetic sequence that has five arithmetic means between 16.2 and 14.7.
A. 16.2, , 15.45, , 14.7
B. 16.2, 16.0, 15.7, 15.5, 15.3, 15.0, 14.7
C. 16.2, 15.95, 15.7, 15.45, 15.2, 14.95, 14.7
D. 16.2, 16.0, 15,8, 15.6, 15,4, 15,2, 15, 14.7
Example 4

27. Find a quadratic model for the sequence 0, 5, 12, 21, 32, 45, … .
Second Differences
Find a quadratic model for the sequence 0, 5, 12, 21, 32, 45, … .
The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an and n into the equation.
0 = a(1)2 + b(1) + c an = 0 and n = 1
5 = a(2)2 + b(2) + c an = 5 and n = 2
12 = a(3)2 + b(3) + c an = 12 and n = 3
This yields a system of linear equations in three variables.
Example 5

28. 0 = a + b + c Simplified first equation
Second Differences
0 = a + b + c Simplified first equation
5 = 4a + 2b + c Simplified second equation
12 = 9a + 3b + c Simplified third equation
Solving for a, b, and c, gives a = 1, b = 2, and c = –3. Substituting these values in the equation for an, the model for the sequence is an = n2 + 2n – 3, as shown in the graph below.
Example 5

29. Second Differences
Answer: an = n2 + 2n – 3
Example 5

30. Find a quadratic model for the sequence 10, 14, 22, 34, 50, … .
A. an = n 2 – 2n + 11
B. an = 2n 2 – 2n + 10
C. an = 4n 2 – 2n + 8
D. an = 10n
Example 5

31. Key Concept 6

32. A. Find the sum of –3 + 2 + 7 + 12 + … + 157.
Sums of Arithmetic Series
A. Find the sum of – …
In this sequence, a1 = –3, an = 157, and d = 2 – (–3) or 5. Use the nth term formula to find the number of terms in the sequence n.
an = a1 + (n – 1)d nth term of an arithmetic sequence
157 = –3 + (n – 1)5 an = 157, a1 = –3, d = 5
33 = n Simplify.
Now use the formula to find the sum of the series.
Example 6

33. Sum of an arithmetic series formula 1.
Sums of Arithmetic Series
Sum of an arithmetic series formula 1.
n = 33, a1 = 3, an = 157
= 2541 Simplify.
Answer: 2541
Example 6

34. Sum of an arithmetic series formula 2.
Sums of Arithmetic Series
B. Find the 17th partial sum of the arithmetic series … .
In this sequence, a1 = 53 and d = 31 – 53 or –22. Use Formula 2 to find the 17th partial sum.
Sum of an arithmetic series formula 2.
n = 17, a1 = 53, d = –22
= –2091 Simplify.
Answer: –2091
Example 6

35. C. = (10 + 6) + (11 + 6) + ··· + (32 + 6) = 16 + 17 + ··· + 38
Sums of Arithmetic Series C.
= (10 + 6) + (11 + 6) + ··· + (32 + 6)
= ··· + 38
The first term of this series is 16, and the last term is 38. The number of terms is equal to the upper bound minus the lower bound plus one, which is 32 – , or 23. Therefore, a1 = 16, an = 38, and n = 23. Use Formula 1 to find the sum of the series.
Example 6

36. Sum of an arithmetic series formula 1.
Sums of Arithmetic Series
Sum of an arithmetic series formula 1.
n = 23, a1 = 16, an = 38
= 621 Simplify.
Answer: 621
Example 6

37. Find
A. 2072
B. 2146
C. 4144
D. 4292
Example 6

38. Sum of Arithmetic Series
MUSIC An artist receives royalties for all CDs of his music that are sold. For the first 10,000 CDs sold, the artist receives $10,000. He receives $12,500 for the next 10,000 CDs, $15,000 for the next 10,000 CDs, and so on. How much money will he receive if 500,000 of his CDs are sold?
First, find the number of terms in the series. Since the artist is paid for every 10,000 CDs, the number of terms is 500,000  10,000 = 50. In this sequence, a1 = 10,000 and d = 12,500 – 10,000 or Use Formula 2 to find the sum of the series.
Example 7

39. He will receive $3,562,500 if 500,000 of his CDs are sold.
Sum of Arithmetic Series
Formula 2
n = 50, a1 =10,000, d = 2500
= 3,562,500 Simplify.
He will receive $3,562,500 if 500,000 of his CDs are sold.
Answer: $3,562,500
Example 7

40. PROFIT The profit a shoe manufacturer makes depends on how many pairs of shoes are sold. For the first 1000 pairs sold, the manufacturer makes $4000 in profit. His profit is $6000 for the next 1000 pairs, $8000 for the next 1000 pairs, and so on. How much profit will the manufacturer make if he sells 20,000 pairs of shoes?
A. $46,000
B. $420,000
C. $460,000
D. $480,000
Example 7

41. Sum of an Arithmetic Series
SALES Anna sells clothing in a store at a mall. Her commission on the sale of new seasonal merchandise is $3.50 per item costing at least $25. To encourage an increase in sales, the store manager promises a $0.05 increase in commission for each additional item sold. Thus, on the sale of the next item, Anna will earn $3.55 in commission. On the sale of the third item, Anna will earn $3.60, and so on. How many items worth $25 or more will Anna need to sell for her total commission on new merchandise to be at least $1000?
Example 8

42. Multiply each side by 2 and simplify.
Sum of an Arithmetic Series
Formula 2
Sn = 1000, a1 =3.50, d = 0.05
Multiply each side by 2 and simplify.
Distribute and subtract 2000 from each side.
Divide each side by 0.05.
Example 8

43. Use the Quadratic Formula.
Sum of an Arithmetic Series
Use the Quadratic Formula.
Simplify.
Because items cannot be negative, Anna will need to sell at least 143 items worth $25 to receive at least $1000 in commission.
Answer: At least 143 items
Example 8

44. Sum of an Arithmetic Series
CHECK
$ $ $ … + $10.60 =
= 
After selling 143 items worth $25 or more, Anna will receive $ in commission, which is more than $1000.
Example 8

45. SALES A toy manufacturer wants to make a profit of $50,000 the first year, $62,000 the second year, $74,000 the third year, and so on. His goal is to make a profit of at least $500,000 per year selling toys. How many years will it take for the manufacturer to reach his goal?
A. 6
B. 7
C. 10
D. 13
Example 8

46. Arithmetic Sequences and Series
LESSON 10–2
Arithmetic Sequences and Series