Name ________________________________________ Date __________________ Class__________________
LESSON
9-3
Reteach
Arithmetic Sequences and Series
To determine whether a sequence is an arithmetic sequence, check for a common
difference, d, d ≠ 0.
Find the first differences of the terms.
2, 6, 18, 54, 162, …
Differences: 6 − 2 = 4
18 − 6 = 12
54 − 18 = 36
162 − 54 = 108
There is no common difference. The
sequence is not arithmetic.
−7, −3, 1, 5, 9, …
Differences: −3 − (−7) = 4
1 − (−3) = 4
5−1=4
9−5=4
The common difference is 4. The
sequence is arithmetic.
If you know the first term of an arithmetic sequence, a1, and the common difference, d, then
you can find the nth term, an , using the following rule.
an = a1 + (n − 1) d
Find the 15th term of the arithmetic sequence 10, 4, −2, −8, −14, …
Step 1
Find the common difference, d.
d = 4 − 10 = −6
Step 2
Identify the first term, a1.
a1 = 10
Step 3
Use the formula with n = 15 to find the 15th term, a15.
an = a1 + (n − 1)d
Write the rule.
a15 = a1 + (15 − 1)d
Substitute n = 15.
a15 = 10 + (14) (−6)
Substitute a1 = 10 and d = −6.
a15 = −74
Simplify.
The 15th term of the sequence is −74.
Determine whether each sequence could be arithmetic. If so, find the
common difference.
1. 3, 15, 27, 39, 51, …
________________________
2. 3, 9, 27, 81, 243, …
_________________________
3. 10, 2, −6, −14, −22, …
________________________
Find the 10th term of each arithmetic sequence.
4. 5, 13, 21, 29, 37, …
5. 7, 4, 1, −2, −5, …
d = _____________
d=
a1 = ______, n = _____
a1 = ______, n = _____
_____________________________________
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9-22
Holt McDougal Algebra 2
Name ________________________________________ Date __________________ Class__________________
LESSON
9-3
Reteach
Arithmetic Sequences and Series (continued)
If you know any two terms in an arithmetic sequence, you can find any other term
in the sequence.
• Find the common difference by using the two terms and the formula for the nth term.
• Then use the formula for the nth term to find the first term and the nth term.
Find the 12th term of the arithmetic sequence with a3 = 33 and a9 = 117.
Step 1
Use the known terms and the formula for the nth term to find the common
difference.
Since an = a9,
an = a1 + (n − 1)d
Write the formula.
n = 9 in
the formula.
a9 = a3 + (9 − 3)d
Let an = a9 and a1 = a3.
Replace 1 with
a9 = a3 + 6d
Simplify.
3 since a1 = a3.
117 = 33 + 6d
Substitute a9 = 117 and a3 = 33.
14 = d
Step 2
Use one of the known terms and the common difference, d = 14, to find a1.
Use a3 = 33 and the formula for the nth term.
an = a1 + (n − 1)d
Write the formula.
a3 = a1 + (3 − 1) (14)
Let an = a3 , so n = 3 and d = 14.
a3 = a1 + (2) (14)
Simplify.
33 = a1 + 28
Substitute a3 = 33.
5 = a1
Step 3
Solve for d.
Solve for a1.
Use a1 = 5, d = 14, and n = 12 in the formula for the nth term to find a12 .
an = a1 + (n − 1)d
Write the formula.
a12 = 5 + (12 − 1) (14)
Substitute a1 = 5, d = 14, and n = 12.
a12 = 5 + (11) (14)
Simplify.
a12 = 159
Solve for a12.
Find the 10th term of the arithmetic sequence with a4 = 34 and a6 = 52.
6. Find d.
7. Find a1.
8. Find a10.
Let an = a6 and a1 = a4.
Let an = a4.
a6 = a4 + (6 − 4) d
a4 = a1 + (4 − 1) (_______)
n = _______
_____________________
________________________
_________________________
________________________
________________________
_________________________
________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-23
Holt McDougal Algebra 2
13. 28.8
14. 12.5
15. 22.6
16. −23.25
+ . . . + n log 10. By Exercise 3, the sum
log10
n ( n + 1) . Since log 10 = 1, this
is
2
n
reduces to ( n + 1) .
2
17. 19 tables
Practice C
1. −30.45
2. 1081
3. −318
4.
103
6
5. 66.4, 47.8, 29.2, 10.6
6. 21.5, 29, 36.5, 44, 51.5
11. −3.24
12. 58.2
13. 816
14. 1266
15. −85
16. 57.5
c. 35; possible answer: the number of
pages Violet must read each day
d. 0 pages; 385 = a1 + (35)(12 − 1); so a1
= 0; and a1 represents the number of
pages for day 1.
Reteach
2. No
3. Yes; d = −8
4. 8; 5; 10; a10 = 77
e. a5 = 0 + (35)(5 − 1); 140 pages
2. a. an = a1 + (n – 1)(2); 20 pages
5. −3; 7; 10; a10 = −20
⎛ 10 + 20 ⎞
b. Sn = 6 ⎜
⎟ ; 90 pages
2
⎝
⎠
6. 52 = 34 + 2d; d = 9
7. 9; 34 = a1 + 27; a1 = 7
3. D
8. 10; a10 = a1 + (10 − 1)d; a10 = 7 + 81;
a10 = 88
1. a. 4
b. Add 4 to each term to get the next
term.
1. a. 2
b. 2n − 1
c. 21
c. n2
2. a. −5
2. a. 3
b. Subtract 5 from each term to get the
next term.
b. 3n
3
n ( n + 1)
2
3. a. k
4. G
Reading Strategy
Challenge
c.
logb a
n ( n + 1)
2
b. Let an = a20 and a1 = a12; replace 1 in (n
− 1) with 12; replace n in (n − 1) with
20; substitute 385 and 665 for a12 and
a20; and calculate d.
17. 1155 candles
1. Yes; d = 12
6.
1. a. an = a1 + (n – 1)d
8. −81.25, −81.5, −81.75
10. −12
ln e
n
n ( n + 1) or ( n + 1)
2
2
Problem Solving
7. −16, −10, −4, 2, 8, 14
9. −30
5.
c. 57
3. a. 23
b. nk
b. Possible answer: I looked at the
numbers and found that the pattern is
to add 12. So I added 12 to 11 to get
23.
k
c. n ( n + 1)
2
4. The series is equivalent to log 10 + log
102 log 103 + . . . + log 10n which by the
Power Property of Logarithms, is
equivalent to log 10 + 2 log 10 + 3 log 10
4. Possible answer: You could use the
pattern and the common difference to find
each term of the sequence up to n = 15.
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A25
Holt McDougal Algebra 2