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- 1. DONE BY, AFSAL M NAHAS XI B KV PATTOM Slide 11-1 Copyright © 2007 Pearson Education, Inc.
- 2. S E QUE NCE A ND S E RIE S Slide 11-2 Copyright © 2007 Pearson Education, Inc.
- 3. Slide 11-3 Copyright © 2007 Pearson Education, Inc.
- 4. Slide 11-4 Copyright © 2007 Pearson Education, Inc.
- 5. Slide 11-5 Copyright © 2007 Pearson Education, Inc.
- 6. Slide 11-6 Copyright © 2007 Pearson Education, Inc.
- 7. Slide 11-7 Copyright © 2007 Pearson Education, Inc.
- 8. An introduction………… 1, 4, 7, 10, 13 35 62 2, 4, 8, 16, 32 9, 1, − 7, − 15 −12 9, − 3, 1, − 1/ 3 20 / 3 6.2, 6.6, 7, 7.4 27.2 85 / 64 1, 1/ 4, 1/16, 1/ 64 π, π + 3, π + 6 3π + 9 π, 2.5π, 6.25π 9.75π Arithmetic Sequences Geometric Sequences ADD MULTIPLY To get next term To get next term Arithmetic Series Geometric Series Sum of Terms Sum of Terms Slide 11-8 Copyright © 2007 Pearson Education, Inc.
- 9. Find the next four terms of –9, -2, 5, … Arithmetic Sequence −2 − −9 = 5 − −2 = 7 7 is referred to as the common difference (d) Common Difference (d) – what we ADD to get next term Next four terms……12, 19, 26, 33 Slide 11-9 Copyright © 2007 Pearson Education, Inc.
- 10. Find the next four terms of 0, 7, 14, … Arithmetic Sequence, d = 7 21, 28, 35, 42 Find the next four terms of x, 2x, 3x, … Arithmetic Sequence, d = x 4x, 5x, 6x, 7x Find the next four terms of 5k, -k, -7k, … Arithmetic Sequence, d = -6k -13k, -19k, -25k, -32k Slide 11-10 Copyright © 2007 Pearson Education, Inc.
- 11. Vocabulary of Sequences (Universal) a1 → First term an → nth term n → number of terms Sn → sum of n terms d → common difference nth term of arithmetic sequence → an = a1 + ( n − 1) d n sum of n terms of arithmetic sequence → Sn = ( a1 + an ) 2 Slide 11-11 Copyright © 2007 Pearson Education, Inc.
- 12. Given an arithmetic sequence with a15 = 38 and d = −3, find a1. a1 → First term x 38 an → nth term n → number of terms 15 NA Sn → sum of n terms d → common difference -3 an = a1 + ( n − 1) d 38 = x + ( 15 − 1) ( −3 ) X = 80 Slide 11-12 Copyright © 2007 Pearson Education, Inc.
- 13. Find S63 of − 19, − 13, −7,... -19 a1 → First term an → nth term 353 ?? n → number of terms 63 Sn → sum of n terms x d → common difference 6 n an = a1 + ( n − 1) d ( a1 + an ) Sn = 2 ?? = −19 + ( 63 − 1) ( 6 ) 63 ( −19 + 353 ) = S63 ?? = 353 2 S63 = 10521 Slide 11-13 Copyright © 2007 Pearson Education, Inc.
- 14. Try this one: Find a16 if a1 = 1.5 and d = 0.5 1.5 a1 → First term an → nth term x n → number of terms 16 NA Sn → sum of n terms d → common difference 0.5 an = a1 + ( n − 1) d a16 = 1.5 + ( 16 − 1) 0.5 a16 = 9 Slide 11-14 Copyright © 2007 Pearson Education, Inc.
- 15. Find n if an = 633, a1 = 9, and d = 24 a1 → First term 9 633 an → nth term n → number of terms x NA Sn → sum of n terms d → common difference 24 an = a1 + ( n − 1) d 633 = 9 + ( x − 1) 24 633 = 9 + 24x − 24 X = 27 Slide 11-15 Copyright © 2007 Pearson Education, Inc.
- 16. Find d if a1 = −6 and a29 = 20 a1 → First term -6 20 an → nth term n → number of terms 29 NA Sn → sum of n terms d → common difference x an = a1 + ( n − 1) d 20 = −6 + ( 29 − 1) x 26 = 28x 13 x= 14 Slide 11-16 Copyright © 2007 Pearson Education, Inc.
- 17. Find two arithmetic means between –4 and 5 -4, ____, ____, 5 a1 → First term -4 an → nth term 5 n → number of terms 4 Sn → sum of n terms NA d → common difference x an = a1 + ( n − 1) d 5 = −4 + ( 4 − 1) ( x ) x=3 The two arithmetic means are –1 and 2, since –4, -1, 2, 5 forms an arithmetic sequence Slide 11-17 Copyright © 2007 Pearson Education, Inc.
- 18. Find three arithmetic means between 1 and 4 1, ____, ____, ____, 4 a1 → First term 1 an → nth term 4 n → number of terms 5 Sn → sum of n terms NA d → common difference x an = a1 + ( n − 1) d 4 = 1 + ( 5 − 1) ( x ) 3 x= 4 The three arithmetic means are 7/4, 10/4, and 13/4 since 1, 7/4, 10/4, 13/4, 4 forms an arithmetic sequence Slide 11-18 Copyright © 2007 Pearson Education, Inc.
- 19. Find n for the series in which a1 = 5, d = 3, Sn = 440 a1 → First term 5 x 440 = ( 5 + 5 + ( x − 1) 3 ) an → nth term y 2 x ( 7 + 3x ) n → number of terms x 440 = 2 440 Sn → sum of n terms 880 = x ( 7 + 3x ) d → common difference 3 0 = 3x 2 + 7x − 880 an = a1 + ( n − 1) d Graph on positive window y = 5 + ( x − 1) 3 X = 16 n Sn = ( a1 + an ) 2 x 440 = ( 5 + y ) 2 Slide 11-19 Copyright © 2007 Pearson Education, Inc.
- 20. Slide 11-20 Copyright © 2007 Pearson Education, Inc.
- 21. An infinite sequence is a function whose domain is the set of positive integers. a1, a2, a3, a4, . . . , an, . . . terms The first three terms of the sequence an = 2n2 are a1 = 2(1)2 = 2 finite sequence a2 = 2(2)2 = 8 a3 = 2(3)2 = 18. Slide 11-21 21 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 22. A sequence is geometric if the ratios of consecutive terms are the same. 2, 8, 32, 128, 512, . . . 8 =4 geometric sequence 2 32 = 4 8 The common ratio, r, is 4. 128 = 4 32 512 = 4 128 Slide 11-22 Copyright © 2007Houghton MifflinInc. Copyright © by Pearson Education, Company, Inc. All rights reserved. 22
- 23. The nth term of a geometric sequence has the form an = a1rn - 1 where r is the common ratio of consecutive terms of the sequence. r = 75 = 5 15 a1 = 15 15, 75, 375, 1875, . . . a2 = a3 = a4 = 15(5) 15(52) 15(53) The nth term is 15(5n-1). Slide 11-23 23 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 24. Example: Find the 9th term of the geometric sequence 7, 21, 63, . . . r = 21 = 3 a1 = 7 7 an = a1rn – 1 = 7(3)n – 1 a9 = 7(3)9 – 1 = 7(3)8 = 7(6561) = 45,927 The 9th term is 45,927. Slide 11-24 24 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 25. The sum of the first n terms of a sequence is represented by summation notation. upper limit of summation n ∑a = a + a + a3 + a4 + L + an i 1 2 i =1 lower limit of summation index of summation 5 ∑ 4n = 41 + 42 + 43 + 44 + 45 n=1 = 4 + 16 + 64 + 256 + 1024 = 1364 Slide 11-25 25 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 26. The sum of a finite geometric sequence is given by ) ( n 1− rn . Sn = ∑ a1r i −1 = a1 1− r i =1 5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ? n=8 r = 10 = 2 a1 = 5 5 )( )( ( )( ) 1 − r n = 5 1 − 28 = 5 1 − 256 = 5 −255 = 1275 Sn = a1 1− 2 −1 1− r 1− 2 Slide 11-26 26 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 27. The sum of the terms of an infinite geometric sequence is called a geometric series. If |r| < 1, then the infinite geometric series a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . . ∞ a1 has the sum S = ∑ a1r = i . 1− r i =0 If r ≥ 1 , then the series does not have a sum. Slide 11-27 27 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 28. Example: Find the sum of 3 − 1 + 1 − 1 +L 39 r = −1 3 a1 = 3 S = () 1− r 1− − 1 3 = 3 = 3 = 3⋅ 3 = 9 1+ 1 4 44 33 The sum of the series is 9 . 4 Slide 11-28 28 Copyright © byPearson Education, Inc. Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 29. 11.3 Geometric Sequences and Series A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant nonzero real number. 1, 2, 4, 8, 16 … is an example of a geometric sequence with first term 1 and each subsequent term is 2 times the term preceding it. The multiplier from each term to the next is called the com onratio and is usually denoted m by r. Slide 11-29 Copyright © 2007 Pearson Education, Inc.
- 30. 11.3 Finding the Common Ratio In a geometric sequence, the common ratio can be found by dividing any term by the term preceding it. The geometric sequence 2, 8, 32, 128, … has common ratio r = 4 since 8 32 128 = = = ... = 4 28 32 Slide 11-30 Copyright © 2007 Pearson Education, Inc.
- 31. 11.3 Geometric Sequences and Series nth Term of a Geometric Sequence In the geometric sequence with first term a1 and common ratio r, the nth term an, is an = a1r n −1 Slide 11-31 Copyright © 2007 Pearson Education, Inc.
- 32. 11.3 Using the Formula for the nth Term ple Find a 5 and a n for the geometric Exam sequence 4, –12, 36, –108 , … Solution Here a 1= 4 and r = 36/–12 = – 3. Using n −1 n=5 in the formula an = a1r 5 −1 a5 = 4 ⋅ (−3) = 4 ⋅ (−3) = 324 4 In general n −1 n −1 an = a1r = 4 ⋅ (−3) Slide 11-32 Copyright © 2007 Pearson Education, Inc.
- 33. 11.3 Modeling a Population of Fruit Flies Exam ple A population of fruit flies grows in such a way that each generation is 1.5 times the previous generation. There were 100 insects in the first generation. How many are in the fourth generation. Solution The populations form a geometric sequence with a1= 100 and r = 1.5 . Using n=4 in the formula for an gives a4 = a1r 3 = 100(1.5)3 = 337.5 or about 338 insects in the fourth generation. Slide 11-33 Copyright © 2007 Pearson Education, Inc.
- 34. 11.3 Geometric Series A g om trics rie is the sum of the terms of a ee es geometric sequence . In the fruit fly population model with a 1 = 100 and r = 1.5, the total population after four generations is a geometric series: a1 + a2 + a3 + a4 = 100 + 100(1.5) + 100(1.5) 2 + 100(1.5)3 ≈ 813 Slide 11-34 Copyright © 2007 Pearson Education, Inc.
- 35. 11.3 Geometric Sequences and Series Sum of the First n Terms of an Geometric Sequence If a geometric sequence has first term a1 and common ratio r, then the sum of the first n terms is given by a1 (1 − r ) n Sn = r ≠1 . where 1− r Slide 11-35 Copyright © 2007 Pearson Education, Inc.
- 36. 11.3 Finding the Sum of the First n Terms 6 ∑ 2 ⋅ 3i Example Find i=1 SolutionThis is the sum of the first six terms of a geometric series with a1 = 2 ⋅ 3 = 6 and r = 3. 1 From the formula for Sn , 6(1 − 3 ) 6(1 − 729) 6(−728) 6 S6 = = = = 2184 . 1− 3 −2 −2 Slide 11-36 Copyright © 2007 Pearson Education, Inc.
- 37. Vocabulary of Sequences (Universal) a1 → First term an → nth term n → number of terms Sn → sum of n terms r → common ratio nth term of geometric sequence → an = a1r n−1 ( ) a1 r n − 1 sum of n terms of geometric sequence → Sn = r −1 Slide 11-37 Copyright © 2007 Pearson Education, Inc.
- 38. Find the next three terms of 2, 3, 9/2, ___, ___, ___ 3 – 2 vs. 9/2 – 3… not arithmetic 3 9/2 3 = = 1.5 → geometric → r = 2 3 2 99 39 3 39 3 3 3 2, 3, , × , × × , × × × 22 22 2 22 2 2 2 9 27 81 243 2, 3, , ,, 2 4 8 16 Slide 11-38 Copyright © 2007 Pearson Education, Inc.
- 39. 1 2 If a1 = , r = , find a9 . 2 3 a1 → First term 1/2 x an → nth term n → number of terms 9 Sn → sum of n terms NA r → common ratio 2/3 an = a1r n−1 9 −1 1 2 x = 2 3 28 27 128 x= 8= 8= 2×3 3 6561 Slide 11-39 Copyright © 2007 Pearson Education, Inc.
- 40. Find two geometric means between –2 and 54 -2, ____, ____, 54 a1 → First term -2 an = a1r n−1 54 an → nth term 54 = ( −2 ) ( x ) 4 −1 n → number of terms 4 −27 = x 3 Sn → sum of n terms NA −3 = x r → common ratio x The two geometric means are 6 and -18, since –2, 6, -18, 54 forms an geometric sequence Slide 11-40 Copyright © 2007 Pearson Education, Inc.
- 41. 2 Find a 2 − a 4 if a1 = −3 and r = 3 -3, ____, ____, ____ 2 Since r = ... 3 −4 −8 −3, − 2, , 39 −8 −10 a 2 − a 4 = −2 − = 9 9 Slide 11-41 Copyright © 2007 Pearson Education, Inc.
- 42. Find a9 of 2, 2, 2 2,... a1 → First term 2 x an → nth term n → number of terms 9 Sn → sum of n terms NA r → common ratio 2 22 r= = =2 2 2 an = a1r n−1 ( 2) 9 −1 x= 2 2 ( 2) 8 x= x = 16 2 Slide 11-42 Copyright © 2007 Pearson Education, Inc.
- 43. If a5 = 32 2 and r = − 2, find a 2 ____, ____, ____,____,32 2 a1 → First term x an → nth term 32 2 n → number of terms 5 Sn → sum of n terms NA −2 r → common ratio an = a1r n−1 () 5 −1 32 2 = x − 2 2 = x( − 2) 4 32 32 2 = 4x 8 2=x Slide 11-43 Copyright © 2007 Pearson Education, Inc.
- 44. *** Insert one geometric mean between ¼ and 4*** *** denotes trick question 1 ,____,4 4 a1 → First term 1/4 an → nth term 4 n → number of terms 3 Sn → sum of n terms NA r → common ratio x an = a1r n−1 1 , 1, 4 4 1 3−1 12 4 = r → 4 = r → 16 = r 2 → ±4 = r 1 4 4 , − 1, 4 4 Slide 11-44 Copyright © 2007 Pearson Education, Inc.
- 45. 111 Find S7 of + + + ... 248 a1 → First term 1/2 an → nth term NA n → number of terms 7 11 Sn → sum of n terms x 1 r= 4 = 8 = r → common ratio 112 ( ) a1 r n − 1 24 Sn = r −1 1 1 7 1 1 7 − 1 − 1 2 2 2 2 63 = x= = 1 1 64 −1 − 2 2 Slide 11-45 Copyright © 2007 Pearson Education, Inc.
- 46. Slide 11-46 Copyright © 2007 Pearson Education, Inc.
- 47. Slide 11-47 Copyright © 2007 Pearson Education, Inc.
- 48. Slide 11-48 Copyright © 2007 Pearson Education, Inc.
- 49. Slide 11-49 Copyright © 2007 Pearson Education, Inc.
- 50. Slide 11-50 Copyright © 2007 Pearson Education, Inc.
- 51. Slide 11-51 Copyright © 2007 Pearson Education, Inc.
- 52. 1, 4, 7, 10, 13, …. No Sum Infinite Arithmetic n Sn = ( a1 + an ) Finite Arithmetic 3, 7, 11, …, 51 2 ( ) a1 r n − 1 Sn = Finite Geometric 1, 2, 4, …, 64 r −1 1, 2, 4, 8, … Infinite Geometric No Sum r>1 r < -1 11 1 a1 Infinite Geometric 3,1, , , ... S= 3 9 27 -1 < r < 1 1− r Slide 11-52 Copyright © 2007 Pearson Education, Inc.
- 53. 111 Find the sum, if possible: 1 + + + + ... 248 11 2 = 4 = 1 → −1 ≤ r ≤ 1 → Yes r= 112 2 a1 1 S= = =2 1 1− r 1− 2 Slide 11-53 Copyright © 2007 Pearson Education, Inc.
- 54. 2 2 + 8 + 16 2 + ... Find the sum, if possible: 8 16 2 = 2 2 → −1 ≤ r ≤ 1 → No r= = 8 22 NO SUM Slide 11-54 Copyright © 2007 Pearson Education, Inc.
- 55. 2111 +++ + ... Find the sum, if possible: 3 3 6 12 11 3 = 6 = 1 → −1 ≤ r ≤ 1 → Yes r= 212 33 2 a1 4 3 S= = = 13 1− r 1− 2 Slide 11-55 Copyright © 2007 Pearson Education, Inc.
- 56. 248 + + + ... Find the sum, if possible: 777 48 r = 7 = 7 = 2 → −1 ≤ r ≤ 1 → No 24 77 NO SUM Slide 11-56 Copyright © 2007 Pearson Education, Inc.
- 57. 5 Find the sum, if possible: 10 + 5 + + ... 2 5 5 2 = 1 → −1 ≤ r ≤ 1 → Yes r= = 10 5 2 a1 10 S= = = 20 1 1− r 1− 2 Slide 11-57 Copyright © 2007 Pearson Education, Inc.
- 58. The Bouncing Ball Problem – Version A A ball is dropped from a height of 50 feet. It rebounds 4/5 of it’s height, and continues this pattern until it stops. How far does the ball travel? 50 40 40 32 32 32/5 32/5 50 40 S= + = 450 4 4 1− 1− 5 5 Slide 11-58 Copyright © 2007 Pearson Education, Inc.
- 59. The Bouncing Ball Problem – Version B A ball is thrown 100 feet into the air. It rebounds 3/4 of it’s height, and continues this pattern until it stops. How far does the ball travel? 100 100 75 75 225/4 225/4 100 100 S= + = 800 3 3 1− 1− 4 4 Slide 11-59 Copyright © 2007 Pearson Education, Inc.
- 60. 11.3 Infinite Geometric Series If a 1, a 2, a 3, … is a geometric sequence and the sequence of sums S1, S2, S3, …is a convergent sequence, converging to a number S∞. Then S∞ is said to be the sum of the infinite geometric series a1 + a2 + a3 + ... = S∞ Slide 11-60 Copyright © 2007 Pearson Education, Inc.
- 61. 11.3 An Infinite Geometric Series Given the infinite geometric sequence 111 1 2, 1, , , , ,... 2 4 8 16 the sequence of sums is S1 = 2, S2 = 3, S3 = 3.5, … The calculator screen shows more sums, approaching a value of 4. So 11 2 + 1 + + + ... = 4 24 Slide 11-61 Copyright © 2007 Pearson Education, Inc.
- 62. 11.3 Infinite Geometric Series Sum of the Terms of an Infinite Geometric Sequence The sum of the terms of an infinite geometric sequence with first term a1 and common ratio r, where –1 < r < 1 is given by a1 S∞ = . 1− r Slide 11-62 Copyright © 2007 Pearson Education, Inc.
- 63. 11.3 Finding Sums of the Terms of Infinite Geometric Sequences i ∞ 3 ple Find ∑ Exam i =1 5 3 3 and r = SolutionHere a1 = so 5 5 3 i ∞ 3 a1 5 =3 ∑ 5 = 1− r = 3 2 . i =1 1− 5 Slide 11-63 Copyright © 2007 Pearson Education, Inc.
- 64. Slide 11-64 Copyright © 2007 Pearson Education, Inc.

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