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# Sequence And Series slayerix

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### Sequence And Series slayerix

1. 1. DONE BY, AFSAL M NAHAS XI B KV PATTOM Slide 11-1 Copyright © 2007 Pearson Education, Inc.
2. 2. S E QUE NCE A ND S E RIE S Slide 11-2 Copyright © 2007 Pearson Education, Inc.
8. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.
21. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.