The document discusses geometric sequences, which are sequences where each term is found by multiplying the previous term by a constant ratio. It provides the recursive formula, explicit formula, examples of finding terms in geometric sequences, and the difference between geometric and arithmetic sequences. Specifically, it notes that geometric sequences use multiplication between terms while arithmetic sequences use addition.

1. Section 7-5
Geometric Sequences

2. Warm-up
Find the next three terms of each sequence.
2. 2, , , ,...
11 1
1. 14, 42, 126, 378, ... 28 32

3. Warm-up
Find the next three terms of each sequence.
2. 2, , , ,...
11 1
1. 14, 42, 126, 378, ... 28 32
1134, 3402, 10206

4. Warm-up
Find the next three terms of each sequence.
2. 2, , , ,...
11 1
1. 14, 42, 126, 378, ... 28 32
, ,
1 1 1
1134, 3402, 10206
128 512 2048

5. Examine the sequence...
48, 72, 108, 162, 243, 364.5, ...
What pattern exists?

6. Examine the sequence...
48, 72, 108, 162, 243, 364.5, ...
What pattern exists?
Multiplying by 1.5 between each term

7. Geometric/Exponential Sequence:

8. Geometric/Exponential Sequence:
A sequence that has first term g1 and a constant
ratio r that you multiply by from term to term

9. Geometric/Exponential Sequence:
A sequence that has first term g1 and a constant
ratio r that you multiply by from term to term
Constant Ratio:

10. Geometric/Exponential Sequence:
A sequence that has first term g1 and a constant
ratio r that you multiply by from term to term
Constant Ratio:
The number you multiply by from term to term;
found by taking one term and dividing by the
previous term; does not equal 0

11. Recursive Formula for a Geometric Sequence
 g1


 gn = rgn−1, for int. n ≥ 2


12. Recursive Formula for a Geometric Sequence
 g1


 gn = rgn−1, for int. n ≥ 2

g1 = first term

13. Recursive Formula for a Geometric Sequence
 g1


 gn = rgn−1, for int. n ≥ 2

g1 = first term
gn = nth term

14. Recursive Formula for a Geometric Sequence
 g1


 gn = rgn−1, for int. n ≥ 2

g1 = first term
gn = nth term
gn -1 = previous term

15. Recursive Formula for a Geometric Sequence
 g1


 gn = rgn−1, for int. n ≥ 2

g1 = first term
gn = nth term
gn -1 = previous term
r = constant ratio

16. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2


17. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 =

18. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 =

19. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) =

20. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12

21. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 =

22. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) =

23. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36

24. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 =

25. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) =

26. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108

27. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 =

28. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) =

29. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324

30. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 =

31. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 = 3(324) =

32. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 = 3(324) = 972

33. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 = 3(324) = 972
g7 =

34. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 = 3(324) = 972
g7 = 3(972) =

35. Example 1
Give the next 6 terms.
 g1 = 4


 gn = 3gn−1, for int. n ≥ 2

g2 = 3g1 = 3(4) = 12
g3 = 3(12) = 36
g4 = 3(36) = 108
g5 = 3(108) = 324
g6 = 3(324) = 972
g7 = 3(972) = 2916

36. Explicit Formula for a Geometric Sequence
n−1
, for int. n ≥ 1
gn = g1r

37. Explicit Formula for a Geometric Sequence
n−1
, for int. n ≥ 1
gn = g1r
g1 = first term

38. Explicit Formula for a Geometric Sequence
n−1
, for int. n ≥ 1
gn = g1r
g1 = first term
gn = th term
n

39. Explicit Formula for a Geometric Sequence
n−1
, for int. n ≥ 1
gn = g1r
g1 = first term
gn = th term
n
r = constant ratio

40. Question
What is the difference between an arithmetic sequence
and a geometric sequence?

41. Question
What is the difference between an arithmetic sequence
and a geometric sequence?
Arithmetic uses addition between terms; geometric uses
multiplication.

42. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1

43. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
g1 = 4(−2)

44. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1 0
g1 = 4(−2) = 4(−2)

45. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)

46. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1
g2 = 4(−2)

47. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1 1
g2 = 4(−2) = 4(−2)

48. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1 1
g2 = 4(−2) = 4(−2) = −8

49. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1 1
g2 = 4(−2) = 4(−2) = −8
3−1 2
g3 = 4(−2) = 4(−2) = 16

50. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1 1
g2 = 4(−2) = 4(−2) = −8
3−1 2
g3 = 4(−2) = 4(−2) = 16
4−1 3
g4 = 4(−2) = 4(−2) = −32

51. Example 2
Write the first 5 terms of the sequence
n−1
gn = 4(−2) , for int. n ≥ 1
1−1
= 4(−2) = 4
0
g1 = 4(−2)
2−1 1
g2 = 4(−2) = 4(−2) = −8
3−1 2
g3 = 4(−2) = 4(−2) = 16
4−1 3
g4 = 4(−2) = 4(−2) = −32
5−1 4
g5 = 4(−2) = 4(−2) = 64

52. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.

53. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r

54. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85

55. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce

56. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce
So what does 6 m represent?

57. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce
So what does 6 m represent?
The initial height (h0)

58. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce
So what does 6 m represent?
The initial height (h0)
h1 = h0 (.85)

59. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce
So what does 6 m represent?
The initial height (h0)
h1 = h0 (.85) = 6(.85)

60. Example 3
A ball is dropped from a height of 6 m and bounces up
to 85% of its previous height after each bounce. Let hn
be the greatest height after the n th bounce. Find a
formula for hn and the height after the 10th bounce.
n−1
, for int. n ≥ 1
hn = h1r
r = .85 h1 = height after first bounce
So what does 6 m represent?
The initial height (h0)
h1 = h0 (.85) = 6(.85) = 5.1

61. Example 3
n−1
, for int. n ≥ 1
hn = h1r

62. Example 3
n−1
, for int. n ≥ 1
hn = h1r
n−1
hn = 5.1(.85) , for int. n ≥ 1

63. Example 3
n−1
, for int. n ≥ 1
hn = h1r
n−1
hn = 5.1(.85) , for int. n ≥ 1
10−1
h10 = 5.1(.85)

64. Example 3
n−1
, for int. n ≥ 1
hn = h1r
n−1
hn = 5.1(.85) , for int. n ≥ 1
10−1 9
h10 = 5.1(.85) = 5.1(.85)

65. Example 3
n−1
, for int. n ≥ 1
hn = h1r
n−1
hn = 5.1(.85) , for int. n ≥ 1
10−1 9
= 5.1(.85) ≈ 1.18 m
h10 = 5.1(.85)

66. Homework

67. Homework
p. 447 #1-23
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