1. Find the common difference of each sequence.
Geometric Sequences
ALGEBRA 1 LESSON 6-4
(For help, go to Lesson 5-6.)
Find the common difference of each sequence.
1. 1, 3, 5, 7, , 17, 15, 13, ...
3. 1.3, 0.1, –1.1, –2.3, , 21.5, 25, 28.5, ...
Use inductive reasoning to find the next two numbers in
each pattern.
5. 2, 4, 8, 16, , 12, 36, ...
7. 0.2, 0.4, 0.8, 1.6, , 100, 50, 25, ...
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2. Common difference: 2 17 – 19 = –2 Common difference: –2
Geometric Sequences
ALGEBRA 1 LESSON 6-4
Solutions
1. 1, 3, 5, 7, , 17, 15, 13, ...
7 – 5 = 2, 5 – 3 = 2, 3 – 1 = – 15 = –2, 15 – 17 = –2,
Common difference: – 19 = –2
Common difference: –2
3. 1.3, 0.1, –1.1, –2.3, , 21.5, 25, 28.5, ...
–2.3 – (–1.1) = –1.2, – – 25 = 3.5, 25 – 21.5 = 3.5,
– 0.1 = –1.2, 0.1 – 1.3 = – – 18 = 3.5
Common difference: –1.2 Common difference: 3.5
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3. Solutions (continued) 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ...
Geometric Sequences
ALGEBRA 1 LESSON 6-4
Solutions (continued)
5. 2, 4, 8, 16, , 12, 36, ...
2(2) = 4, 4(2) = 8, 8(2) = 16, 4(3) = 12, 12(3) = 36,
16(2) = 32, 32(2) = (3) = 108, 108(3) = 324
Next two numbers: 32, 64 Next two numbers: 108, 324
7. 0.2, 0.4, 0.8, 1.6, , 100, 50, 25, ...
(0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 200  2 = 100, 100  2 = 50,
1.6, 1.6(2) = 3.2, 3.2(2) =  2 = 25, 25  2 = 12.5,
Next two numbers: 3.2,  2 = 6.25
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4. (–5) (–5) (–5)  Geometric Sequences
ALGEBRA 1 LESSON 6-4
Find the common ratio of each sequence.
a. 3, –15, 75, –375, . . .
3 –15 75 –375
(–5) (–5) (–5)
The common ratio is –5.
b. 3, 3 2 4 8 ,
, ... 3 2 4 8  1
The common ratio is . 1 2
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5. Find the next three terms of the sequence 5, –10, 20, –40, . . .
Geometric Sequences
ALGEBRA 1 LESSON 6-4
Find the next three terms of the sequence 5, –10, 20, –40, . . .
5 –10 20 –40
(–2) (–2) (–2)
The common ratio is –2.
The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.
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6.  Geometric Sequences Determine whether each sequence is arithmetic
ALGEBRA 1 LESSON 6-4
Determine whether each sequence is arithmetic
or geometric.
a. 162, 54, 18, 6, . . . 1 3 
The sequence has a common ratio.
The sequence is geometric.
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7. The sequence has a common difference.
Geometric Sequences
ALGEBRA 1 LESSON 6-4
(continued)
b. 98, 101, 104, 107, . . .
+ 3
The sequence has a common difference.
The sequence is arithmetic.
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8. first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
Geometric Sequences
ALGEBRA 1 LESSON 6-4
Find the first, fifth, and tenth terms of the sequence that has the rule A(n) = –3(2)n – 1.
first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48
tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536
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9. The first term is 2 meters, which is 200 cm.
Geometric Sequences
ALGEBRA 1 LESSON 6-4
Suppose you drop a tennis ball from a height of 2 meters. On each bounce, the ball reaches a height that is 75% of its previous height. Write a rule for the height the ball reaches on each bounce. In centimeters, what height will the ball reach on its third bounce?
The first term is 2 meters, which is 200 cm.
Draw a diagram to help understand the problem.
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10. A rule for the sequence is A(n) = 200 • 0.75n – 1.
Geometric Sequences
ALGEBRA 1 LESSON 6-4
(continued)
The ball drops from an initial height, for which there is no bounce. The initial height is 200 cm, when n = 1. The third bounce is n = 4. The common ratio is 75%, or 0.75.
A rule for the sequence is A(n) = 200 • 0.75n – 1.
A(n) = 200 • 0.75n – 1
Use the sequence to find the height of the third bounce.
A(4) = 200 • – 1
Substitute 4 for n to find the height of the third bounce.
= 200 • 0.753
Simplify exponents.
= 200 •
Evaluate powers. =
Simplify.
The height of the third bounce is cm.
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11. 2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
Geometric Sequences
ALGEBRA 1 LESSON 6-4
1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . .
2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
3. Determine whether each sequence is arithmetic or geometric.
a. 37, 34, 31, 28, . . .
b. 8, –4, 2, –1, . . .
4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1.
5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so that its dimensions are 20% larger than its original size. Write a rule for the length of the copies. What will be the length if you enlarge the photograph five times? (Hint: The common ratio is not just 0.2. You must add 20% to 100%.) –2
3, 1, 1 3
arithmetic
geometric
4, 2500, 1,562,500
A(n) = 6(1.2)n-1; about 14.9 in.
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