M11PC - UNIT REVIEW: Series & Sequences

Name: Class: Date: ID: A M11PC - UNIT REVIEW: Series & Sequences Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The first three terms of the sequence defined by t n = 0.3n + 0.5 are A 0.5, 0.8, 1.1 C 0., 0.1, 0.4 B 0.3, 0., 0.7 D 0.3, 0.8, 1.3 Completion Complete each statement. 1. In the formula for the general term of an arithmetic sequence t n = a + ( n 1)d, a represents the and n represents the.. 9 + 17 + 5 + 33 + 41 + ë is an example of a(n). 3. The geometric sequence 0, 60, 180, 540,, 393 660 has terms. 4. The sum of the first 8 terms of the geometric series 5 15 + 45 135 + ë is. 5. The ratio used to generate the infinite geometric series.5 + 0.5 0.1 + 0.0 + ë is. Matching Match the correct term to its description below. A arithmetic series D divergent series B geometric sequence E geometric series C arithmetic sequence F infinite geometric series 1. a geometric series that has no last term. the sum of the terms of a sequence in which the terms have a common ratio 3. the sum of the terms of a sequence in which the terms have a common difference 4. a sequence where there is a common difference between consecutive terms 5. a sequence where there is a common ratio between consecutive terms 1

Name: ID: A Short Answer For each arithmetic sequence, determine a) the value of and d b) an explicit formula for the general term c) t 0 1. 8, 5,, 1,. 3a, 3a b, 3a 4b, 3a 6b, 3. 7 4, 1, 1 4, 1, 4. The starting wage at a bookstore is $8.50 per hour with a yearly increase of $0.75 per hour. a) Write the general term of the sequence representing the hourly rate earned in each year. b) Use your expression from part a) to determine the hourly rate after 6 years. c) How many years will someone need to work at the store to earn $15.5 per hour? For each geometric sequence, determine a) an explicit formula for the general term b) 1 5. = 3, r = 6. 3,, 4 3, 8 9, 16 7,... 7. 3, 3 3, 9, 9 3, 7,... 8., 5, 5, 15,... For each arithmetic series, determine a) an explicit formula for the general term b) a formula for the general sum c) d) S n 9. =, d = 3, n = 4 10. + 4 + 6 + ë + 48 11. 1 9 6 ë + 1

Name: ID: A Determine the sum of each arithmetic series. 1. = 3 3, d = 3, n = 11 13. 4k + 11k + 18k +ë + 74k 14. ( 4a 3b) + ( 4a + b) + ( 4a + 5b) +ë + ( 4a + 9b) 15. Find the value of given S 8 = 380 and r = 3. Be sure to show all of your work. 16. If S 1 = 0.7 and S =.1 in a geometric series, determine the sum of the firs terms in the series. Be sure to show all of your work. 17. A bouncy ball bounces to its height when it is dropped on a hard surface. Suppose the ball is dropped from 3 0 m. a) What height will the ball bounce back up to after the sixth bounce? b) What is the total distance the ball travels if it bounces indefinitely? Problem 1. Consider an arithmetic sequence with t = 1 3 and t 8 = 13 3. a) Determine the values of and d. b) Determine an explicit formula for the general term of the sequence. c) Determine the value of each term. i) ii) t 0 iii) t 6 d) Determine the term number of each term. i) 7 3 ii) 9 iii) 97 3. In a lottery to join a golf club, the first person drawn from the names must pay $14 000. Each subsequent person drawn pays $50 less than the person before. The last person drawn pays $8000 for a membership. a) Write the first four terms of the sequence that represents the cost of a membership. b) Determine and d for the sequence. c) Determine an explicit formula for the general term. d) What will the 10th golfer pay for a membership? e) How many golfers will be able to join the club?

Name: ID: A 3. At the end of the second week after opening, a new fitness club has 870 members. At the end of the seventh week, there are 1110 members. Suppose the increase is arithmetic. a) How many members joined the club each week? b) How many members were there in the first week? 4. The sum of the first two terms of an arithmetic series is 15 and the sum of the next two terms is 43. What are the first four terms of the series? 5. Etienne owns a small recycling company that picks up empty glass bottles from restaurants. At the first restaurant, he picks up 50 bottles. At each restaurant after this, he picks up 4 more bottles than he picked up at the restaurant before. Assume that this pattern continues. a) Write the first four terms of the arithmetic sequence that represents the number of bottles he picks up at a restaurant. b) Determine a and d for the sequence. c) How many bottles will he have picked up after stopping at the eighth restaurant? d) If his truck can hold 000 empty bottles, will he be able to pick up bottles at the 1st restaurant without emptying the truck first? 6. In an arithmetic series, the sum of the first 5 terms is 70 and the sum of the firs1 terms is 35. a) Determine the values of and d. b) What is the sum of the first i) 30 terms? ii) 50 terms? iii) 100 terms? c) Describe how you could use the formula for the sum of an arithmetic series to find the total of terms 35 to 40. d) Use your method from part c) to determine the sum of terms 35 to 40. 7. A toy car is rolling down an inclined track and picking up speed as it goes. The car travels 4 cm in the first second, 8 cm in the second second, 1 cm in the next second, and so on. What is the total distance travelled by the car in the first 30 s? 8. A company purchases a new computer system valued at $4 000. For income tax purposes, an accountant determines that the annual depreciation rate (rate of decrease in value) for the equipment is 11%. a) Make a table of values to show the value of the system over the first 5 years. b) Determine an explicit formula in function notation to model the value of the system in year n. c) What is the value of the system at the end of year 0? d) How realistic is the answer to part c)? Explain. 9. For each sequence, determine i) an explicit formula for the nth term, using function notation ii) f(11) a) 1, 3, 3, 3 3, 9, 9 3, b) 5, 13, 5, 41, 61, c) 1 7, 1 3, 5 11, 7 13, 3 5,... 4

Name: ID: A 10. The value of an antique increases in value at a rate of.5% every year. In 000, the antique was purchased for $5000. a) Determine an explicit formula to represent the value of the antique since the year 000. b) Use your formula to write the first three terms of the sequence. c) What was the value of the antique in 008? d) In which year will the value of the antique be $11 866? 11. For a geometric series, S 4 = 1. What are the first three terms of the series if the first term is 3? S 8 17 1. Write each repeating decimal number as an equivalent fraction in lowest terms. a) 0.5555... b) 0.1 5

M11PC - UNIT REVIEW: Series & Sequences Answer Section MULTIPLE CHOICE 1. ANS: C PTS: 1 DIF: Easy OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: terms explicit formula arithmetic sequence COMPLETION 1. ANS: a = first term n = term number d = common difference PTS: 1 DIF: Easy OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: general term arithmetic sequence. ANS: arithmetic series PTS: 1 DIF: Easy OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: arithmetic series 3. ANS: 10 PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF 10 TOP: Geometric Sequences KEY: number of terms geometric sequence 4. ANS: 800 PTS: 1 DIF: Average OBJ: Section 1.4 NAT: RF 10 TOP: Geometric Series KEY: sum geometric series 5. ANS: 0. PTS: 1 DIF: Easy OBJ: Section 1.5 NAT: RF 10 TOP: Infinite Geometric Series KEY: ratio MATCHING 1. ANS: F PTS: 1 DIF: Easy OBJ: Section 1.5 NAT: RF 10 TOP: Infinite Geometric Series KEY: infinite geometric series. ANS: E PTS: 1 DIF: Easy OBJ: Section 1.4 NAT: RF 10 TOP: Geometric Series KEY: geometric series 3. ANS: A PTS: 1 DIF: Easy OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: arithmetic series 4. ANS: C PTS: 1 DIF: Easy OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: arithmetic sequence 1

5. ANS: B PTS: 1 DIF: Easy OBJ: Section 1.3 NAT: RF 10 TOP: Geometric Sequences KEY: geometric sequence SHORT ANSWER 1. ANS: a) = 8, d = 3 b) t n = 8 + ( n 1)(3) = 8 + 3n 3 = 3n 11 c) t 0 = 3( 0) 11 = 60 11 = 49 PTS: 1 DIF: Easy OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: terms explicit formula arithmetic sequence. ANS: a) = 3a, d = b b) t n = 3a + ( n 1) ( b) = 3a bn + b c) t 0 = 3a b( 0) + b = 3a 40b + b = 3a 38b PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: terms explicit formula arithmetic sequence

3. ANS: a) = 7 4, d = 3 4 b) t n = 7 4 + ( n 1) Ê 3 ˆ Á 4 = 7 4 3 4 n + 3 4 = 3 4 n + 5 c) t 0 = 3 4 ( 0) + 5 = 15 + 5 = 5 PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: terms explicit formula arithmetic sequence 4. ANS: a) t n = 8.50 + (n 1)(0.75) = 8.50 + 0.75n 0.75 = 7.75 + 0.75n b) t 6 = 7.75 + 0.75( 6) = 1.5 The hourly rate after 6 years is $1.5. c) 15.5 = 7.75 + 0.75n 7.50 = 0.75n n = 10 You would need to work at the bookstore for 10 years to earn $15.5 per hour. PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: explicit formula terms 5. ANS: a) t n = 3() n 1 b) 1 = 3() 11 1 = 3() 10 = 307 PTS: 1 DIF: Easy OBJ: Section 1.3 NAT: RF 9 TOP: Geometric Sequences KEY: explicit formula terms geometric sequence 3

6. ANS: a) Since = 3 and r = 3, t n = 3 n 1 Ê ˆ Á 3 Ê b) 1 = 3 ˆ Á 3 = 3 10 Ê ˆ Á 3 = 104 19 683 11 1 PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF 9 TOP: Geometric Sequences KEY: explicit formula terms geometric sequence 7. ANS: a) Since = 3 and r = 3, t n = 3( 3) n 1 b) 1 = 3( 3) 11 1 = 3( 3) 10 = 79 PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF 9 TOP: Geometric Sequences KEY: explicit formula terms geometric sequence 8. ANS: a) Since = and r = 1 5, t n = 1 n 1 Ê ˆ Á 5 Ê b) 1 = 1 ˆ Á 5 = 1 10 Ê ˆ Á 5 = 9 765 65 11 1 PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF 9 TOP: Geometric Sequences KEY: explicit formula terms geometric sequence 4

9. ANS: a) t n + (n 1)d = + (n 1)(3) = 3n 1 b) S n = n [ + (n 1)d] = n [() + (n 1)(3)] = n ( 3n + 1) c) = 3( 1) 1 = 35 d) S 4 = 4 (3(4) + 1) = 6 PTS: 1 DIF: Easy OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: explicit formula sum terms arithmetic series 10. ANS: a) t n + (n 1)d = + (n 1)() = n b) S n = n [ + (n 1)d] = n È ÎÍ () + (n 1)() = n ( n + ) = n + n c) = (1) = 4 d) Since t n = 48, 48 = n n = 4 S 4 = 4 + 4 = 600 PTS: 1 DIF: Average OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: explicit formula sum terms arithmetic series 5

11. ANS: a) t n + (n 1)d = 1 + (n 1)(3) = 3n 15 b) S n = n [ + (n 1)d] = n [( 1) + (n 1)(3)] = n (3n 7) c) = 3( 1) 15 = 1 d) Since t n = 1, 1 = 3n 15 3n = 7 n = 9 S 9 = 9 (3(9) 7) = 0 PTS: 1 DIF: Average OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: explicit formula sum terms arithmetic series 1. ANS: S n = n [ + (n 1)d] S 11 = 11 = 11 = 11 = 77 3 [(3 3) + ( 10)( 3)] (6 3 0 3) ( 14 3) PTS: 1 DIF: Easy OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: sum arithmetic series 6

13. ANS: = 4k, d = 7k t n + (n 1)d = 4k + (n 1)(7k) = 4k + 7kn 7k = 7kn 3k t n = 74k 74k = 7kn 3k 77k = 7kn n = 11 S n = n [ + (n 1)d] S 11 = 11 = 11 = 49k [(4k) + ( 10 )(7k)] ( 8k + 70k ) PTS: 1 DIF: Average OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: sum arithmetic series 7

14. ANS: = 4a 3b, d = 4b t n + (n 1)d = (4a 3b) + (n 1)(4b) = 4a 3b + 4nb 4b = 4a 7b + 4nb t n = 4a + 9b 4a + 9b = 4a 7b + 4nb 36b = 4nb n = 9 S n = n [ + (n 1)d] S 9 = 9 [(4a 3b) + (8)(4b)] = 9 ( 8a 6b + 3b ) = 9 (8a + 6b) = 9( 4a + 13b) = 36a + 117b PTS: 1 DIF: Difficult OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: sum arithmetic series 15. ANS: S n (rn 1) r 1 380 [( 3)8 1] 3 1 380 = t ( 6561 1) 1 4 13 10 ( 6560) = PTS: 1 DIF: Average OBJ: Section 1.4 NAT: RF 10 TOP: Geometric Series KEY: sum geometric series 8

16. ANS: In the series, = 0.7. To find t, evaluate S S 1. t = S S 1 =.1 0.7 = 1.4 To find r, evaluate t. r = t = 1.4 0.7 = So, r =. S n (r n 1) r 1 S 1 = 0.7(1 1) 1 = 0.7(4096 1) 1 = 866.5 PTS: 1 DIF: Difficult OBJ: Section 1.4 NAT: RF 10 TOP: Geometric Series KEY: sum geometric series 9

17. ANS: a) Draw a diagram of the situation. Ê = 0 ˆ and r = Á 3 3 = 40 3 t n r n 1 t 6 = 40 3 Ê ˆ Á 3 5 1.76 The ball will bounce to a height of approximately 1.76 m on the sixth bounce. b) The distance the ball travels (both down and up) is 0 + + t + t 3 + ë = 0 + S. 10

Ê t ˆ 1 0 + S = 0 + 1 r Á Ê 40 ˆ 3 = 0 + 1 3 Á Ê ˆ = 0 + Á Ê 40 3 1 3 ˆÊ = 0 + 40 3ˆ Á 3 Á 1 = 0 + 80 = 100 The total distance the ball will travel is 100 m. PTS: 1 DIF: Difficult OBJ: Section 1.5 NAT: RF 10 TOP: Infinite Geometric Series KEY: sum infinite geometric series 11

PROBLEM 1. ANS: a) Use t n = a + (n 1)d to write and solve a system of equations. t + ( 1)d and t 8 + (8 1)d 1 3 = t + d 1 (1) 13 3 + 7d () 1 3 + d (1) 1 3 = 6d () (1) 4 = 6d 13 3 + 7d () d = 3 Substitute d = into one of the original equations in the system. 3 1 3 + 3 = 1 3 b) t n + (n 1)d = 1 3 + ( n 1) Ê ˆ Á 3 = 1 3 + 3 n 3 = 3 n 1 1

c) i) = 3 (1) 1 ii) t 0 = 3 (0) 1 iii) t 6 = 3 (6) 1 d) i) = 8 1 = 7 7 3 = 3 n 1 = 40 3 1 = 37 3 ii) 9 = 3 n 1 iii) = 5 3 1 = 49 3 97 3 = 3 n 1 10 3 = 3 n 9 + 1 = 3 n 100 3 = 3 n n = 5 10 = 3 n n = 50 n = 15 PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: explicit formula arithmetic sequence terms. ANS: a) 14 000, 13 750, 13 500, 13 50 b) = 14 000, d = 50 c) t n + (n 1)d = 14 000 + ( n 1) ( 50) = 14 000 50n + 50 = 14 50 50n d) 0 = 14 50 50( 10) = 11 750 Therefore, the 10th golfer will pay $11 750. e) The last person pays $8000, so t n = 8000. t n = 14 50 50n 8000 = 14 50 50n 650 = 50n n = 5 Therefore, 5 golfers will be able to join the club. PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: arithmetic sequence explicit formula terms 13

3. ANS: a) t n + (n 1)d t = 870 t 7 = 1110 t 7 t = (7 1)d ( 1)d = 5d 1110 870 = 5d 40 = 5d d = 48 Each week, 48 members joined the club. b) t + d 870 + 48 = 8 There were 8 members in the first week. PTS: 1 DIF: Average OBJ: Section 1.1 NAT: RF 9 TOP: Arithmetic Sequences KEY: arithmetic sequence explicit formula terms 4. ANS: + t = 15 t 3 + t 4 = 43 + ( + d) = 15 ( + d) + ( + 3d) = 43 + d = 15 + 5d = 43 Solve the system of equations to determine and d. + d = 15 + 5d = 43 4d = 8 d = 7 Substitute d = 7 into the first equation to solve for. + d = 15 + 7 = 15 = 8 = 4 The first four terms are 4, 11, 18, and 5. PTS: 1 DIF: Difficult OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: arithmetic series explicit formula terms 14

5. ANS: a) 50, 54, 58, 6 b) = 50, d = 4 c) S n = n [ + (n 1)d] S 8 = 8 [(50) + (8 1)(4)] = 4(100 + 8) = 51 He will have picked up 51 bottles after stopping at the eighth restaurant. d) S 0 = 0 [ ( 50 ) + ( 0 1) ( 4)] = 10(100 + 76) = 1760 Use the explicit formula for the general term of an arithmetic sequence to determine the number of bottles at the 1st restaurant. t n + (n 1)d = 50 + ( 1 1) ( 4) = 50 + ( 0) ( 4) = 50 + 80 = 130 He has 1760 bottles in the truck and will be picking up 130 more at the next restaurant. This would bring the total to 1760 + 130, or 1890, which is less than 000. This means that he can stop at the next restaurant. PTS: 1 DIF: Difficult OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: arithmetic sequence terms sum arithmetic series 15

6. ANS: a) Use S n = n [ + (n 1)d] to write and solve a system of equations. S 5 = 5 [ + (5 1)d] 70 = 5 [ + 4d] 70 = 5 + 10d 14 + d Solve the system of equations. 3 + 5d 14 + d 18 = 3d d = 6 Substitute d = 6 into one of the equations. 14 + d 14 + (6) = =, d = 6 b) S n = n [ + (n 1)d] S 11 = 11 [ + (11 1)d] 35 = 11 [ + 10d] 35 = 11 + 55d 3 + 5d = n [( ) + ( n 1)6] = n (4 + 6n 6) = n ( 6n ) = n( 3n 1) i) S 30 = 30(3( 30) 1) = 30( 90 1) = 30( 89) ii) S 50 = 50(3( 50) 1) = 50( 150 1) = 50( 149) iii) S 100 = 100(3( 100) 1) = 100( 300 1) = 100( 99) = 670 = 7450 = 9 900 c) Find the sum of the first 34 terms and the sum of the first 40 terms and subtract the two results. d) S 40 = 40(3( 40) 1) and S 34 = 34(3( 34) 1) = 40( 10 1) = 40( 119) = 34( 10 1) = 34( 101) = 4760 = 3434 Therefore, the required sum is 4760 3434, or 136. 16

PTS: 1 DIF: Difficult + OBJ: Section 1.1 Section 1. NAT: RF 9 TOP: Arithmetic Sequences Arithmetic Series KEY: sum arithmetic series 7. ANS: = 4 and d = 4. S n = n [t + ( n 1)d] 1 S 30 = 30 [( 4) + ( 30 1)4] = 15[8 + ( 9)4] = 15[ 14] = 1860 The toy car travels 1860 cm or 18.6 m. PTS: 1 DIF: Easy OBJ: Section 1. NAT: RF 9 TOP: Arithmetic Series KEY: sum arithmetic series 8. ANS: a) Year Value ($) 0 4 000.00 1 37 380.00 33 68.0 3 9 608.70 4 6 351.74 5 3 453.05 b) V(n) = 4 000(0.89) n, where V represents the value and n represents the number of years since the system was purchased. c) V(0) = 4 000(0.89) 0 =Ö 4083.66 The value of the system after 0 years is $4083.66. d) This value is most likely not realistic, as most companies would replace a computer system before the 0-year mark to upgrade the system to current needs. This would often be done before the end of the 10th year of owning a new system (or less, for some companies that require more up-to-date computer technology). PTS: 1 DIF: Difficult + OBJ: Section 1.3 NAT: RF 10 TOP: Geometric Sequences KEY: model geometric sequence explicit formula 17

9. ANS: a) i) f(n) = f(1)r n 1 = ( 3) n 1 ii) f(11) = ( 3) 11 1 = ( 3) 10 = 43 b) i) Graph the sequence on a coordinate grid with n on the horizontal axis and t n on the vertical axis. The points look like they lie on a parabola. Thus, the formula for the nth term is of the form f(n) = an + bn + c. Substitute f(1), f(), and f(3) into this formula and solve for a, b, and c. 5 = a(1) + b(1) + c 13 = a() + b() + c 5 = a(3) + b(3) + c 5 = a + b + c (1) 13 = 4a + b + c () 5 = 9a + 3b + c (3) Subtract (1) from (): Subtract () from (3): 13 = 4a + b + c 5 = 9a + 3b + c 5 = a + b + c 13 = 4a + b + c 8 = 3a + b (4) Subtract (4) from (5): 1 = 5a + b 8 = 3a + b 1 = 5a + b (5) 4 = a a = Substituting a = into 8 = 3a + b gives b =. Substituting a = and b = into a + b + c = 5 gives c = 1. Thus, the formula for the nth term is f(n) = n + n + 1. ii) f(11) = (11) + (11) + 1 = 4 + + 1 = 65 18

c) i) The denominators of, t 3, and t 4 look like part of the sequence 7, 9, 11, 13, 15,... Rewriting t with a denominator of 9 gives 3 9, and rewriting t 5 with a denominator of 15 gives 9. Thus, the sequence is 15 1 7, 3 9, 5 11, 7 13, 9 15,. The numerator is an arithmetic sequence with a = 1 and d =, and the denominator is an arithmetic sequence with a = 7 and d =. Thus, 1 + (n 1) f(n) = 7 + (n 1) = 1 + n 7 + n = n 1 n + 5 ii) f(11) = ( 11) 1 ( 11) + 5 = 1 + 5 = 1 7 = 7 9 PTS: 1 DIF: Difficult + OBJ: Section 1.3 NAT: RF 10 TOP: Geometric Sequences KEY: terms explicit formula sequence 10. ANS: a) At a rate of increase of.5% per year, r = 1.05. The initial value of the antique in 000 is $5000, so t 0 = 5000. t n = 5000(1.05) n b) = 5000( 1.05) 1 t = 5000( 1.05) t 3 = 5000( 1.05) 3 = 515 553.13 5384.45 The first three terms of the sequence are $515, $553.13, and $5384.45. c) The year 008 represents t 8. t 8 = 5000( 1.05) 8 =Ö 609.01 d) 11 866 = 5000( 1.05) n.373 = ( 1.05) n Using systematic trial, ( 1.05) 35.373. So, n = 35. In 035, the antique will be worth approximately $11 866. PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF 9 TOP: Geometric Sequences KEY: explicit formula geometric sequence terms 19

11. ANS: S 4 S 8 = 1 17 (r 4 1) r 1 (r 8 1) r 1 = 1 17 r 4 1 r 8 1 = 1 17 17r 4 17 = r 8 1 r 8 17r 4 + 16 = 0 (r 4 1)(r 4 16) = 0 r 4 = 1 or r 4 = 16 r = ±1 or r = ± From the definition of a geometric series, r ±1, so the first three terms are 3, 6, 1 or 3, 6, 1. PTS: 1 DIF: Difficult OBJ: Section 1.4 NAT: RF 10 TOP: Geometric Series KEY: explicit formula sum geometric series 0

1. ANS: a) 0. 5 = 0.5 + 0.05 + 0.005 +ë This is an infinite geometric series with = 0.5 and r = 0.1. S 1 r = 0.5 1 0.1 = 0.5 0.9 = 5 9 b) 0.1 = 0.1 + 0.0 + 0.00 + 0.000 +ë After 0.1, = 0.0 and r = 0.1. S 1 r = 0.0 1 0.1 = 0.0 0.9 = 90 Add this fraction to 0.1 (or 1 10 ): 1 10 + 90 = 9 + 90 = 11 90 PTS: 1 DIF: Difficult OBJ: Section 1.5 NAT: RF 10 TOP: Infinite Geometric Series KEY: explicit formula sum infinite geometric series 1

M11PC - UNIT REVIEW: Series Sequences [Answer Strip] ID: A C 1. F 1. E. A 3. C 4. B 5.