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2.3 Sequences & Series

2.3 Sequences & Series. 2 Credits - Internal. Sequences. A sequence is a succession of numbers separated usually by commas. e.g. 5, 7, 9, 11, … or 8, 16, 32, 64, … Each member of the sequence is called a term and may be represented by the letter t. t 1

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2.3 Sequences & Series

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  1. 2.3 Sequences & Series 2 Credits - Internal

  2. Sequences A sequence is a succession of numbers separated usually by commas. e.g. 5, 7, 9, 11, … or 8, 16, 32, 64, … Each member of the sequence is called a term and may be represented by the letter t t1 (1st Term) t2 (2nd Term)

  3. Sequences The nth term tells us the general rule for the sequence. e.g. 5, 7, 9, 11, … or 8, 16, 32, 64, … 2 2 2 tn= 2n + 3 tn= 4 x 2n

  4. Sequences Write down the next 3 terms of the following sequences. a.) 5, 9, 13, 17, 21, … b.) 3, -6, 12, -24, … c.) 2, 5, 10, 17, 26 … d.) 0, 1, 1, 2, 3, 5, 8, 13, … 25, 29, 33 tn= 4n + 1 4 4 4 4 tn= 3 x (-2)n-1 48, -96, 192 common ratio -2 37, 50, 65 tn= n2 + 1 3 5 7 9 2 2 2 n2 = 1, 4, 9, 16, 25 21, 34, 55 Fibonacci

  5. Sequences Find the first three terms of each sequence. a.)tn= 5n + 3 b.)tn= n3 – n2 c.)tn= (3n – 2)2 8, 13, 18 0, 4, 18 1, 16, 49

  6. Arithmetic Sequences Arithmetic sequences have a common difference (d) between successive terms. (linear) t1 t2t3 t4 a + 3d a a + 1d a + 2d The general rule for the nth term of an arithmetic sequence tn = a + (n – 1)d a = first term d = common difference

  7. Arithmetic Sequences Find the formula for the nth term and the 20th term of the arithmetic sequence 5, 9, 13, 17, 21, … a = d = 5 4 tn = a + (n – 1)d tn = 5 + (n – 1)4 = 4n + 1 t20 = 4(20) + 1 = 81

  8. Arithmetic Sequences Find the number of terms in the arithmetic sequence 23, 29, 35, 41, 47, …, 125 tn = a + (n – 1)d a = d = 23 6 tn = 23 + (n – 1)6 = 6n + 17 125 = 6n + 17 There are 18 terms in the sequence 108 = 6n n = 18

  9. tn = a + (n – 1)d Starter Find the formula for the nth term and the 20th term of each of these arithmetic sequences 47, 66, 85, 104, … 11, 4, -3, -10, … a = d = 11 -7 a = d = 47 19 tn = 11 + (n – 1)(-7) tn = 47 + (n – 1)19 = 19n + 28 = 18 – 7n t20 = -7(20) + 18 t20 = 19(20) + 28 = -122 = 408

  10. tn = a + (n – 1)d Starter Find the number of terms in each of these arithmetic sequences 11, 14, 17, 20, …, 116 5, 7, 9, 11, 13, …, 1957 a = d = 11 3 a = d = 5 2 tn = 11 + (n – 1)3 tn = 5 + (n – 1)2 = 2n + 3 = 3n + 8 116 = 3n + 8 1957 = 2n + 3 1954 = 2n 108 = 3n n = 977 n = 36

  11. Arithmetic Sequences The second term of an arithmetic sequence is 28 and the 6th term is 0. Find the first four terms of the sequence t1 t2t3 t4 28 28 + d 28 + 2d a t6 = 0 28 + 4d = 0 The first four terms are: 35, 28, 21, 14, … 4d = -28 d = -7

  12. Arithmetic Sequences An arithmetic sequence has a 5th term of 24 and a 7th term of 40. What are the first three terms and the rule for the sequence. Solution:Between the 5th and 7th terms we have 2 common differences. 2d = 40 – 24 2d = 16 d = 8 tn = a + (n – 1)8 tn = -8 + (n – 1)8 24 = a + (5 – 1)8 tn = 8n – 16 24 = a + 32 The first 3 terms are -8, 0, 8 -8 = a

  13. Try These! Find the first four terms of the arithmetic sequences with given terms. a.)t1 = 6 and t8 = 34 b.) t1 = 12 and t15 = 124 c.) t6 = 15 and t12 = 27 d.) t5 = 60 and t16 = -105 6, 10, 14, 18 12, 20, 28, 36 5, 7, 9, 11 120, 105, 90, 75,

  14. tn = a + (n – 1)d Merit Paddy is an aspiring tennis player and his coach asks him to follow a coaching program. The total time Paddy must practice each week increases by the same amount. In week five, the program required him to practice 410 minutes and in week eight he has to practice for 470 minutes. In what week will Paddy spend over 12 hours practicing? (720 minutes) t5 = 410 410 = a + (5 – 1)(20) t8 = 470 410 = a + (4)(20) Between term 5 and term 8 there are 3 common differences a = 330 720 3d = 470 – 410 tn = 330 + (n – 1)20 d = 20 n = 20.5 (the 21st week)

  15. tn = a + (n – 1)d Merit A plumber’s time is charged at an initial call out charge of a dollars and a constant rate of d dollars for each minute he works. One job that takes 35 minutes is charged at $52.25. Another that takes one and a half hours (90 min) is charged at $93.50 What is the formula that relates the length of the job to the cost? 52.25 = a + (35 – 1)d 52.25 = a + 34d 93.50 = a + 89d 93.50 = a + (90 – 1)d -41.25 = -55d d = 0.75 52.25 = a + (35 – 1)(0.75) tn = $26.75 + (n – 1)($0.75) a = 26.75 tn = $0.75n + $26

  16. tn = a + (n – 1)d Merit A plumber’s time is charged at an initial call out charge of a dollars and a constant rate of d dollars for each minute he works. One job that takes 35 minutes is charged at $52.25. Another that takes one and a half hours (90 min) is charged at $93.50 How much would he charge for a two hour job? tn = $0.75n + $26 = $0.75(120) + $26 = $116.00

  17. Sums of Arithmetic Sequences The letter S denotes a partial sum and the number subscript identifies the number of terms being summed Sn = Sn = n/2 (a + L) last term first term

  18. Sn= Starter Find the sum of the first 30 terms of the arithmetic sequence 6, 10, 14, 18,… a = d = n = 6 4 30 Sn = = 15 [ 2(6) + (29)(4)] = 15 x 128 = 1920 150 Find the sum of the first 25 terms of 54, 50, 46, 42,…

  19. Sn= Starter Find the sum of the following arithmetic sequence 247, 229, 211, 193, … , -581 tn = a + (n – 1)8 a = d = n = 247 -18 -581 = 247 + (n – 1)-18 n = 47 47 Sn = = 23.5 [ 2(247) + (46)(-18)] = 23.5 x -334 = -7849 Find the sum of the following sequence 1, 2, 3, 4 ,…, 1 000 000 500 000 500 000

  20. Sigma Notation The symbol Σ means “the sum of” and is the Greek capital letter sigma. The numbers below and above the sigma are the first and the last terms to be added. Example: 20 ∑ 2n + 3 1

  21. Sigma Notation To calculate the answer: • Find the first term by substituting the first number into the formula • Find the second term by substituting the next number into the formula • Continue until you reach the last term • Add the terms e.g. = 2 + 6 + 10 + 14 + 18 t1 = 4(1) –2 = 2 t4 = 4(4) –2 = 14 t2 = 4(2) –2 = 6 = 50 t5= 4(5) –2 = 18 t3= 4(3) –2 = 10

  22. Sigma Notation e.g. Write the following in sigma notation • 5, 8, 11, 14 • 4 + 8 + 12 + 16 + 20 = 31+2 + 32+2 + 33+2 + 34+2 3n + 2 = = 14 + 24 + 34 + 44 + 54 4n = THETA Pg 96 Ex 12.3

  23. Sn= Starter After how many terms does the sum of 8, 14, 20, 26, 32, …. equal 572? a = d = n = 8 6 Sn = ? 572 = n/2 [2(8) + (n – 1)(6)] 1144 = n [10 + 6n] 0 = 6n2 + 10n - 1144 0 = 3n2 + 5n - 572 n = 13 orn = -14.67

  24. Sn= Starter After how many terms does the sum of 68, 60, 52, 44, …. equal 308? a = d = n = 68 -8 Sn = ? 308 = n/2 [2(68) + (n – 1)(-8)] 616 = n [144 – 8n] 0 = -8n2 + 144n – 616 0 = n2 – 18n + 77 n = 7 or n = 11

  25. Sn= Try These After how many terms does the sum of 9, 4, -1, -6, -11, … equal -1905? 13, 21, 29, 37, … equal 10 045? a = d = n = 9 -5 Sn = -1905 = n/2 [2(9) + (n – 1)(-5)] ? -3810 = n [23 – 5n] 0 = -5n2 + 23n + 3810 n = -25.4 or n = 30 THETA Pg 107 Ex 13.2 13 8 a = d = n = 49

  26. Sn= Merit After how many terms does the sum of the arithmetic sequence 7, 12, 17, 22, 27, … equal 4599? 22, 18, 14, 10, ….. equal 70? (no calculators) a = d = n = 7 5 Sn = 4599 = n/2 [2(7) + (n – 1)(5)] ? 9198 = n [9 + 5n] 0 = 5n2 + 9n – 9198 n = -43.8 or n = 42 n = 7 or n = 5

  27. Graphing Sequences A sequence can be represented by graphing the value of the term against the value of n. The points are not joined, because sequences exist for natural numbers only. tn = ½ × 2ntn= 3n – 2 THETA Pg 97 Ex 12.4

  28. Applications From the information given we have to decide if we are using the term or summing formula. e.g. Jane who is preparing to run a marathon (42km) decides that the best way to get fit is to run a little further each day until she is capable of running the marathon distance. Her running schedule is as follows: Day 1 – 1 km , Day 2 – 1.5 km, Day 3 – 2 km, etc • a.) How far will Jane run on Day 10 of her schedule? tn = a + (n – 1)d t10 = 1 + (10 – 1) x 0.5 = 5.5 km • b.) On what day of her schedule will Jane finally run the marathon distance of 42 km? 42 = 1 + (n – 1) x 0.5 n = 83 days

  29. Applications From the information given we have to decide if we are using the term or summing formula. e.g. Jane who is preparing to run a marathon (42km) decides that the best way to get fit is to run a little further each day until she is capable of running the marathon distance. Her running schedule is as follows: Day 1 – 1 km , Day 2 – 1.5 km, Day 3 – 2 km, etc c.) By the time Jane runs the marathon distance found in (b) on the 83rd day, what is the total distance she will run altogether? S83 = (2x1 + (83 - 1) x 0.5) THETA Pg 109 Ex 13.3 = 1784.5 km

  30. Applications e.g. A wholesaler sells packets of dried soup to supermarkets. The more packets a supermarket purchases the cheaper the cost per packet and consequently the greater profit they can make. • The price structure is as follows: • for 1000 packets of dried soup the cost per packet is 30 cents • for 2000 packets the cost per packet is 29.5 cents • for 3000 packets the cost per packet is 29 cents... • (this pricing pattern continues) a.) Represent the selling cost per 1000’s of soup packets as a sequence of numbers (listing the first 6 terms) 2000 3000 4000 5000 6000 30¢ 29.5¢ 29¢ 28.5¢ 28¢ 27.5¢

  31. Applications e.g. A wholesaler sells packets of dried soup to supermarkets. The more packets a supermarket purchases the cheaper the cost per packet and consequently the greater profit they can make. b.) What is the cost per packet of dried soup if a supermarket purchases 5000 packets? T5 = 28¢ c.) What is the total cost for the purchase of 6000 packets of soup? T6 = 27.5¢ ea. 6000 x 0.275 = $1650 d.) A supermarket chain purchases 20 000 packets. What is their unit cost per packet and the total cost of the purchase? tn = a + (n – 1)d T20= 30 + (20 – 1)(-0.5) = 20.5 ¢ ea. 20 000 x 0.205 = $4100

  32. Applications e.g. Tradesmen usually buy different spanners so they have the right tool for the job. The shop assistant noted that the spanners from company A were heavy with the stated mass of the first five as 115 g, 135 g, 155 g, 175 g, and 195 g and they continue to increase by the same amount. What does the full set of the first 20 spanners weigh? Sn= S20 = 10 [2(115) + (19)(20)] = 6100 g (6.1 kg)

  33. Merit e.g. As part if an advertising promotion a car dealer chooses a car from his lot with a retail price of $23 000 and each day that the car is not sold he reduces the price by $400 a.) If the car is sold on Day 12 how much is it sold for? tn = a + (n – 1)d T12 = 23 000 + (11)(-400) = $18 600 b.) On what day can the car be purchased for $15 800? 15 800 = 23 000 + (n – 1)(-400) 15 800 = -400n + 23 400 n = 19

  34. Merit e.g. As part if an advertising promotion a car dealer chooses a car from his lot with a retail price of $23 000 and each day that the car is not sold he reduces the price by $400 a.) The car cost the dealer $12 500. After how many days will the dealer be losing money on the vehicle? tn = a + (n – 1)d 12 500 = -400n + 23 400 n = 27.25 After 28 days the dealer will lose $$$ b.) If no one purchases the car, after how many days will it be selling for $0? 0 = -400n + 23 400 n = 58.5 After 59 days its free !

  35. Excellence e.g. Ann has been saving for her retirement. She has saved $89 640 after 24 years. Each year she has increased her savings from the previous year by the same amount. In year 10 she saved $3210. How much money did Ann save in the first year of her savings plan? tn = a + (n – 1)d Sn= 3210 = a + (9)d a = 3210 – 9d 89 640 = 12[2(3210 – 9d) + 23d] a = 3210 – 9(210) 89 640 = 12[6420+ 5d] a = $1320 (savings in first year) 89 640 = 77040 + 60d d = 210

  36. Geometric Sequences • In a geometric sequence, each term is calculated by multiplying the previous term by the same number each time. • This number is called the common ratio. • If the sequence is decreasing, the ratio is going to be a fraction e.g. 2, 8, 32, 128, … a sequence with a common ratio of 36, -12, 4, , … a sequence with a common ratio of 4 tn = a  rn-1 General Formula: • where a is the first term • r is the common ratio found by r = • tnis the nth term

  37. tn = a  rn-1 e.g. Find the first 4 terms in the geometric sequence where a = 3 and r = 4. t1 = 3  40 = 3 3, 12, 48, 192, … t2 = 3  41 = 12 t3 = 3  42 = 48 t3 = 3  43 = 192 You Try! Find the first 4 terms in the geometric sequence where a = 5 and r = 2 5, 10, 20, 40, …

  38. tn = a  rn-1 e.g. What is the common ratio (r) in the following sequence? 64, 16, 4, … r = = = You Try! What is the common ratio (r) in the following sequence? 6, -30, 150, -750, ... r = r = -5

  39. tn = a  rn-1 e.g. The first four terms in a sequence are 4, 8, 16, 32, ... Calculate the 7th term. t7 = 4  27-1 a = r = n = 4 r = = 256 2 = 2 THETA Pg 114 Ex 14.1 7 You Try! Find the formula for the nth term of the geometric sequence 4, 12, 36, 108, 324, ... as well as the 20th term 4 a = r = n = tn = 4  3n-1 r = 3 t20 = 4 649 045 868 20 r = 3

  40. tn = a  rn-1 Starter The 4th term of a geometric sequence is -72 and the 7th term is 576. Find the first 3 terms. …., -72, ___, ___, 576, … a = r = ? …., -72, -72r, -72r2, -72r3, … -2 -72r3 = 576 r3 = -8 r = -2 Dividing by -2 we get the first three terms of the sequence 9, -18, 36, -72, …

  41. tn = a  rn-1 Try these The 1st term of a geometric sequence is 7 and the 4th term is 189. Find the first 3 terms. 7 a = r = 7, ___, ___, 189, … 3 7, 7r, 7r2, 7r3, … <7, 21, 63, …> 7r3 = 189 r3 =27 r = 3 The 3rd term of a geometric sequence is 256 and the 6th term is 2048. Find the first 3 terms. r = 2 <64, 128, 256, …>

  42. tn = a  rn-1 Example Find the number of terms in the geometric sequence 2, 4, 8, 16, …, 8192. 2 a = r = 8192 = 2 × 2n – 1 2 4096 = 2n – 1 log4096 = log2n – 1 log(An) = n log A log4096 = (n – 1)log2 log4096 = n – 1 log2 n = 13

  43. tn = a  rn-1 Example Find the number of terms in the geometric sequence 65 536, 32 768, 16 384, … , 16. 65 536 a = r = tn = a  rn-1 0.5 16 = 65 536 × (0.5)n – 1 0.0002441406 = (0.5)n – 1 log.0002441406 = log(0.5)n – 1 log(An) = n log A log.0002441406 = (n – 1)log(0.5) log.0002441406 = n – 1 log(0.5) n = 13

  44. tn = a  rn-1 Try These Find the number of terms in the geometric sequence 11, 33, 99, 297, … , 157 837 977. 11 a = r = tn = a  rn-1 3 157 837 977 = 11 × (3)n – 1 THETA Pg 114 Ex 14.1 14348907 = (3)n – 1 log 14348907 = log(3)n – 1 log(An) = n log A log 14348907 = (n – 1)log(3) log 14348907 = n – 1 log 3 n = 16 Find the number of terms in the geometric sequence 16, 12, 9, 27/4, … , 2187/1024 r = 0.75, n = 8

  45. Applications e.g. On the first day that Matt starts for the soccer season he runs 4.00 km. The distance he runs each day increases by 15% of the distance he ran the day before. If Matt keeps to his programme, what is the distance he will run on the 12th day? r = 1.15 a = 4.00 tn = a  rn-1 a = r = r = 1.15 t12 = 4.00  (1.15)12-1 t12 = 18.61 km

  46. Applications e.g. Susie suspends a 21.0 g copper sulphate crystal in a copper sulphate solution. Each day the crystal increases its mass by 18 % from the previous day. Susie intends to leave the crystal in the solution for 20 days. What will be the mass of the crystal then? r = 1.18 tn = a  rn-1 a = r = 21.0 1.18 t20 = 21.0  (1.18)20-1 t20 = 487.5 g

  47. tn = a  rn-1 Starter The 5th term of a geometric sequence is a5b4and the 8th term is a8b7. Find the first 3 terms. …., a5b4, ____ , ____ , a8b7 , … a = r = ? ….,, a5b4, a5b4 r, a5b4 r2, a5b4 r3… ab a5b4r3 = a8b7 r3 = a3b3 r = ab Dividing by ab we get the first three terms of the sequence a, a2b , a3b2 , ….

  48. Sum of the first n terms in a Geometric Sequence e.g.Calculate the sum (to the nearest whole number) of the first 9 terms of the geometric sequence 125, 225, 405, 729, … Sn = 125(1.89 – 1) 1.8 – 1 a = r = n = 125 = 1.8 THETA Pg 115 Ex 14.2 Sn = 30837 9

  49. tn = a  rn-1 Starter Find the number of terms in the geometric sequence 800, 200, 50, 12.5, … , 0.1953125 800 a = r = tn = a  rn-1 0.25 0.1953125 = 800 × (0.25)n – 1 0.0002441406 = (0.25)n – 1 log.0002441406 = log(0.25)n – 1 log(An) = n log A log.0002441406 = (n – 1)log(0.25) log.0002441406 = n – 1 log(0.25) n = 7

  50. Applications e.g. A computer artificial intelligence system is designed to learn from its own mistakes. When it was trialled on understanding some written English, it made 850 errors. Every trial after this it makes only 95% of the errors of the previous trial. After how many trials will the number of errors drop to 100 or less? r = 0.95 tn = a  rn-1 a = r = 850 0.95 tn = 850  (0.95)n-1 100 > 850  (0.95)n-1 0.117647 > (0.95)n-1 log.117647 > log(0.95)n-1 n = 43 log.117647 > (n–1)log(0.95) 41.7 > (n–1)

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