AA Section 7-5

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Geometric Sequences

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  • AA Section 7-5

    1. 1. Section 7-5 Geometric Sequences
    2. 2. Warm-up Find the next three terms of each sequence. 2. 2, , , ,... 11 1 1. 14, 42, 126, 378, ... 28 32
    3. 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. 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. 5. Examine the sequence... 48, 72, 108, 162, 243, 364.5, ... What pattern exists?
    6. 6. Examine the sequence... 48, 72, 108, 162, 243, 364.5, ... What pattern exists? Multiplying by 1.5 between each term
    7. 7. Geometric/Exponential Sequence:
    8. 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. 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. 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. 11. Recursive Formula for a Geometric Sequence  g1    gn = rgn−1, for int. n ≥ 2 
    12. 12. Recursive Formula for a Geometric Sequence  g1    gn = rgn−1, for int. n ≥ 2  g1 = first term
    13. 13. Recursive Formula for a Geometric Sequence  g1    gn = rgn−1, for int. n ≥ 2  g1 = first term gn = nth term
    14. 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. 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. 16. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2 
    17. 17. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2  g2 =
    18. 18. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2  g2 = 3g1 =
    19. 19. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2  g2 = 3g1 = 3(4) =
    20. 20. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2  g2 = 3g1 = 3(4) = 12
    21. 21. Example 1 Give the next 6 terms.  g1 = 4    gn = 3gn−1, for int. n ≥ 2  g2 = 3g1 = 3(4) = 12 g3 =
    22. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 36. Explicit Formula for a Geometric Sequence n−1 , for int. n ≥ 1 gn = g1r
    37. 37. Explicit Formula for a Geometric Sequence n−1 , for int. n ≥ 1 gn = g1r g1 = first term
    38. 38. Explicit Formula for a Geometric Sequence n−1 , for int. n ≥ 1 gn = g1r g1 = first term gn = th term n
    39. 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. 40. Question What is the difference between an arithmetic sequence and a geometric sequence?
    41. 41. Question What is the difference between an arithmetic sequence and a geometric sequence? Arithmetic uses addition between terms; geometric uses multiplication.
    42. 42. Example 2 Write the first 5 terms of the sequence n−1 gn = 4(−2) , for int. n ≥ 1
    43. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 61. Example 3 n−1 , for int. n ≥ 1 hn = h1r
    62. 62. Example 3 n−1 , for int. n ≥ 1 hn = h1r n−1 hn = 5.1(.85) , for int. n ≥ 1
    63. 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. 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. 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. 66. Homework
    67. 67. Homework p. 447 #1-23 “It was on my fifth birthday that Papa put his hand on my shoulder and said, ‘Remember, my son, if you ever need a helping hand, you’ll find one at the end of your arm.’” - Sam Levenson
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