Chapter 1 Sequences and Series, Summaries of Pre-Calculus

Download Chapter 1 Sequences and Series and more Summaries Pre-Calculus in PDF only on Docsity! MHR • Pre-Calculus 11 Solutions Chapter 1 Page 1 of 80 Chapter 1 Sequences and Series Section 1.1 Arithmetic Sequences Section 1.1 Page 16 Question 1 a) 32 − 16 = 16, 48 − 32 = 16, 64 − 48 = 16, … Since successive differences are constant, the sequence is arithmetic. t1 = 16; d = 16. The next three terms are: 80 + 16 = 96 96 + 16 = 112 112 + 16 = 128 b) Not arithmetic, because the differences of consecutive terms are not constant: 4 − 2 = 2, 8 − 4 = 4,… c) −7 − (−4) = −3, −10 − (−7) = −3, −13 − (−10) = −3, … Since successive differences are constant, the sequence is arithmetic. t1 = −4; d = −3. The next three terms are: −16 + (−3) = −19 −19 + (−3) = −22 −22 + (−3) = −25 d) 0 − 3 = −3, −3 − 0 = −3, −6 − (−3) = −3, … Since successive differences are constant, the sequence is arithmetic. t1 = 3; d = −3. The next three terms are: −9 + (−3) = −12 −12 + (−3) = −15 −15 + (−3) = −18 Section 1.1 Page 16 Question 2 a) t1 = 5, t2 = 5 + 3 or 8, t3 = 8 + 3 or 11, t4 = 11 + 3 = 14 The first four terms are 5, 8, 11, 14. b) t1 = −1, t2 = −1 + (−4) or −5, t3 = −5 + (−4) or −9, t4 = −9 + (−4) or −13 The first four terms are −1, −5, −9, −13. c) t1 = 4, t2 = 4 + 1 5 or 14 5 , t3 = 1 14 5 5 + or 24 5 , t4 = 2 14 5 5 + or 34 5 The first four terms are 4, 14 5 , 24 5 , 34 5 . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 2 of 80 d) t1 = 1.25, t2 = 1.25 + (−0.25) or 1, t3 = 1 + (−0.25) or 0.75, t4 = 0.75 + (−0.25) = 0.5 The first four terms are 1.25, 1, 0.75, 0.5. Section 1.1 Page 16 Question 3 a) tn = 3n + 8 t1 = 3(1) + 8 t1 = 11 b) t7 = 3(7) + 8 t7 = 29 c) t14 = 3(14) + 8 t14 = 50 Section 1.1 Page 16 Question 4 a) t4 = 19, t5 = 23 t5 − t4 = 23 − 19 or 4 d = 4 tn = t1 + (n − 1)d 19 = t1 + (4 − 1)4 19 = t1 + 12 t1 = 7 Then, t2 = 7 + 4 or 11, and t3 = 11 + 4 or 15. The missing terms are 7, 11, 15; t1 = 7 and d = 4. b) t3 = 3, t4 = 3 2 t4 − t3 = 3 2 − 3 or 3 2 − d = 3 2 − tn = t1 + (n − 1)d 3 = t1 + (3 − 1) 3 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ − 3 = t1 + (−3) t1 = 6 Then, t2 = 6 + 3 2 ⎛ ⎞−⎜ ⎟ ⎝ ⎠ or 14 2 . The missing terms are 6, 14 2 ; t1 = 6 and d = 3 2 − . c) t2 = 4, t5 = 10 t5 = t2 + 3d 10 = 4 + 3d 6 = 3d d = 2 The missing terms are 2 and 6, 8; t1 = 2 and d = 2. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 5 of 80 8 5slope 2 1 slope 3 − = − = The slope is the same as the common difference. It is the coefficient of the variable term in the general term, 2 + 3n. e) If a line were drawn through the points, the y-intercept would be 2. This is the same as the constant value in the general term, 2 + 3n. Section 1.1 Page 17 Question 8 Consider A: tn = 6 + (n − 1)4 34 = 6 + 4n − 4 34 − 2 = 4n 32 = 4n n = 8 The sequence defined by tn = 6 + (n − 1)4 has 34 as its 8th term. Consider B: tn = 3n − 1 34 = 3n − 1 35 = 3n This does not give a natural number for n, the term number, so 34 cannot be a term of this sequence. Consider C: t1 = 12, d = 5.5 Then, tn = t1 + (n − 1)d tn = 12 + (n − 1)5.5 34 = 12 + 5.5n − 5.5 34 − 6.5 = 5.5n 27.5 = 5.5n 27.5 5.5 5 n n = = The sequence for which t1 = 12 and d = 5.5 has 34 as its fifth term. Consider D: From the pattern of the sequence, t1 = 3 and d = 4. Then, tn = 3 + (n − 1)4 34 = 3 + 4n − 4 34 + 1 = 4n This does not give a natural number for n, the term number, so 34 cannot be a term of this sequence. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 6 of 80 Section 1.1 Page 17 Question 9 t16 = 110, d = 7 Substitute for t16 in tn = t1 + (n − 1)d. 110 = t1 + 15(7) 110 = t1 + 105 t1 = 5 The first term of the sequence is 5. Section 1.1 Page 17 Question 10 t1 = 5y, d = −3y Substitute into tn = t1 + (n − 1)d tn = 5y + (n − 1)(−3y) tn = 5y − 3ny + 3y tn = 8y − 3ny For t15, substitute n = 15. t15 = 8y − 3(15)y t15 = 8y − 45y t15 = −37y Section 1.1 Page 17 Question 11 The difference between consecutive terms is the same for terms of an arithmetic sequence. 7x − 4 − (5x + 2) = 10x + 6 − (7x − 4) 7x − 4 − 5x − 2 = 10x + 6 − 7x + 4 2x − 6 = 3x + 10 x = −16 Substitute to find the three terms: 5x + 2 = 5(−16) + 2 7x − 4 = 7(−16) − 4 10x + 6 = 10(−16) + 6 = −78 = −116 = −154 Check that these terms have a common difference: −116 − (−78) = −38 −154 − (−116) = −38 So, the value of x is −16 and the three terms are −78, −116, −154. Section 1.1 Page 17 Question 12 The difference between consecutive terms is the same for terms of an arithmetic sequence. z − y = y − x z = 2y − x MHR • Pre-Calculus 11 Solutions Chapter 1 Page 7 of 80 Section 1.1 Page 17 Question 13 a) Perimeters of the four figures shown: 10, 16, 22, 28. The perimeters are an arithmetic sequence with t1 = 10 and d = 6. So, an equation for the perimeter of figure n is Pn = 10 + (n − 1)6 or Pn = 6n + 4. b) For the perimeter of Figure 9, substitute n = 9. P9 = 10 + (9 − 1)6 P9 = 58 The perimeter of Figure 9 is 58 units. c) Determine the value of n when Pn = 76. 76 = 10 + (n − 1)6 76 = 10 + 6n − 6 72 = 6n n = 12 Figure 12 has a perimeter of 76 units. Section 1.1 Page 17 Question 14 a) t1 = 0, d = 8 The tee-off times are: 8:00, 8:08, 8:16, 8:24, … Considering 8:00 to be time 0, the sequence is 0, 8, 16, 24. b) Extend the sequence to 60 min. 0, 8, 16, 24, 32, 40, 48, 56, … So within the first hour, 8 groups of four will have teed-off. This means 32 players will be on the course after 1 h. c) tn = 0 + (n − 1)8 tn = 8n − 8 d) For 132 players, there will need to be 132 ÷ 4 or 33 groups teeing-off. Substitute n = 33 into tn. tn = 8(33) − 8 tn = 256 This means that the last group will tee-off 256 min after the first group. 256 min = 4 h 16 min, so the last group will tee-off at 8:00 + 4:16 or 12:16. e) Answers may vary. Examples: Rain may interrupt the tee-off times or players in a group might not be quite ready at their tee-off time. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 10 of 80 Section 1.1 Page 19 Question 19 a) Depth (ft) 0 30 60 90 Water Pressure (psi) 14.7 29.4 44.1 58.8 The first four terms of the sequence of water pressure with depth are 14.7, 29.4, 44.1, 58.8. The general term of this sequence is tn = 14.7n, where n is the number of 30-ft descents. b) First determine n for 1000 ft. 1000 133 30 3 = Water pressure 114.7 33 1 14.7 3 490 ⎛ ⎞= + −⎜ ⎟ ⎝ ⎠ = The pressure at a depth of 1000 ft is 490 psi. Determine n for 2000 ft. 2000 266 30 3 = Water pressure 214.7 66 1 14.7 3 980 ⎛ ⎞= + −⎜ ⎟ ⎝ ⎠ = The pressure at a depth of 1000 ft is 980 psi. c) d) The y-intercept is 14.7. e) The slope is 14.7. f) The y-intercept is t1 and the slope is the common difference. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 11 of 80 Section 1.1 Page 19 Question 20 Let a represent the length of the shortest side and d the common difference. Then, the four sides are: t1, t1 + d, t1 + 2d, and t1 + 3d. Use the perimeter: t1 + t1 + d + t1 + 2d + t1 + 3d = 60 4t1 + 6d = 60 2t1 + 3d = 30 Given that the longest side is 24 cm, t1 + 3d = 24 Subtract from . t1 = 6 Substitute t1 = 6 in (2) to find d. 6 + 3d = 24 3d = 18 d = 6 The other three sides lengths are 6 cm, 12 cm and 18 cm. Check: 6 + 12 + 18 + 24 = 60. Section 1.1 Page 19 Question 21 a) Term Number 1 2 3 4 5 Number of minutes 4 8 12 16 20 Number of degrees 1 2 3 4 5 The sequence of number of minutes is 4, 8, 12, 16, 20, … b) The number of minutes, tn, is given by tn = 4n where n is the number of degrees of turn. c) The time for a rotation of 80º is 4(80) or 320 min. This is 5 h 20 min. Section 1.1 Page 19 Question 22 Year 1986 1987 … 2006 2007 Term Number 1 22 Number of Beekeepers 1657 1048 t1 = 1657, tn = 1048, n = 22 tn = t1 + (n − 1)d 1048 = 1657 + (21)d −609 = 21d d = −29 From 1986 to 2007, the number of beekeepers decreased by about 29 each year. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 12 of 80 Section 1.1 Page 19 Question 23 Year 2003 2004 … 2022 2023 Term Number 1 20 Millions of Carats 3.8 113.2 t1 = 3.8, tn = 113.2, n = 20 tn = t1 + (n − 1)d 113.2 = 3.8 + 19d 109.4 = 19d d ≈ 5.76 The common difference is approximately 5.8 million carats. This is the increase in the number of diamond carats extracted each year between 2003 and 2023. Section 1.1 Page 20 Question 24 The radius for the circle traversed by wheel 12 will be 50 + 11(20) or 270 m. Use C = πd. C = π(270)(2) C ≈ 1696.460… Wheel 12 traverses a circle with circumference of about 1696.5 m. Section 1.1 Page 21 Question 25 a) Consider each time to be number of minutes after 13:00. Then, the first five terms are 54, 59, 64, 69, 74; t1 = 54, d = 5 b) tn = t1 + (n − 1)d tn = 54 + (n − 1)5 tn = 5n + 49 c) The terms are numbers of minutes after 13:00. d) Determine the time when n = 24. t24 = 5(24) + 49 t24 = 169 So, 169 min after 13:00 the sun was completely eclipsed. This is 2 h 49 min after 13:00, so the complete eclipse occurred at 15:49. Section 1.1 Page 21 Question 26 a) An arithmetic sequence is an increasing sequence if and only the difference between one term and the next is a positive number. An arithmetic sequence is an increasing sequence if and only if d > 0. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 15 of 80 c) t1 = 8, d = −5, tn = −102 Step 1: Determine n. Substitute into tn = t1 + (n − 1)d. −102 = 8 + (n − 1)(−5) −102 = 8 − 5n + 5 5n = 115 n = 23 Step 2: Determine the sum of the series. Substitute into 1( ) 2n n nS t t= + . 2 23 3 23 8 102( ( )) 2 1081 S S = + = − − d) t1 = 2 3 , d = 1, tn = 41 3 Step 1: Determine n. Substitute into tn = t1 + (n − 1)d. ( 1) 1 41 2 1 3 3 3 1 14 n n n = + − + = = Step 2: Determine the sum of the series. Substitute into 1( ) 2n n nS t t= + . 4 14 1 2 301 1 or 100 3 3 14 2 41 3 3 S S ⎛ ⎞= +⎜ ⎟ ⎝ ⎠ = Section 1.2 Page 27 Question 2 a) t1 = 1, d = 2, n = 8 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 8 8 8 8 1 8 2[2( ) ( 1) ] 2 4[2 14] 64 S S S = + − = + = MHR • Pre-Calculus 11 Solutions Chapter 1 Page 16 of 80 b) t1 = 40, d = −5, n = 11 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 1 11 1 11 [2( ) ( 1)( )] 2 11[8 11 4 0 50] 2 165 0 11 5S S S = + − = − = − c) t1 = 1 2 , d = 1, n = 7 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 8 7 8 7 1 72 ( 1) 2 7 [1 6] 2 49 or 24.5 2 1 2 S S S ⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ = + = d) t1 = −3.5, d = 2.25, n = 6 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 6 6 6 [2( ) ( 1) ] 2 3[ 7 6 3.5 6 11.25] 12.75 2.25S S S = + − = − + = − Section 1.2 Page 27 Question 3 a) t1 = 7, tn = 79, n = 8 Substitute into 1( ) 2n n nS t t= + . 8 8 8 8 7 79( ) 2 4(86) 344 S S S = + = = MHR • Pre-Calculus 11 Solutions Chapter 1 Page 17 of 80 b) t1 = 58, tn = −7, n = 26 Substitute into 1( ) 2n n nS t t= + . 2 26 6 26 26 58 7( ( )) 2 13(51) 663 S S S = = −= + c) t1 = −12, tn = 51, n = 10 Substitute into 1( ) 2n n nS t t= + . 1 10 0 10 10 12 5( ) 2 5 1 (39) 195 S S S + = = −= d) t1 = 12, d = 8, n = 9 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 9 9 9 [2( ) ( 1)( )] 2 9 [24 64] 2 9 12 9 9 8 3 6 S S S = + − = + = e) t1 = 42, d = −5, n = 14 Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 14 1 14 4 14 42 14 5[2( ) ( 1)( )] 2 7[84 65] 133 S S S = + − = − = − MHR • Pre-Calculus 11 Solutions Chapter 1 Page 20 of 80 b) t1 = 10, d = −3, n = 10 For t10, substitute into tn = t1 + (n − 1)d. t10 = 10 + (10 − 1)(−3) t10 = −17 For S10, substitute into 1[2 ( 1) ] 2n nS t n d= + − . 10 1 10 0 10 10 10 3[2( ) ( 1)( )] 2 5[20 27] 35 S S S = + − = − = − − c) t1 = −10, d = −4, n = 10 For t10, substitute into tn = t1 + (n − 1)d. t10 = −10 + (10 − 1)(−4) t10 = −10 − 36 t10 = −46 For S10, substitute into 1[2 ( 1) ] 2n nS t n d= + − . 10 1 10 0 10 10 10 4[2( ) ( 1)( )] 2 5[ 20 36] 280 S S S = + − = − − = − − − d) t1 = 2.5, d = 0.5, n = 10 For t10, substitute into tn = t1 + (n − 1)d. t10 = 2.5 + (10 − 1)(0.5) t10 = 7 For S10, substitute into 1[2 ( 1) ] 2n nS t n d= + − . 10 10 10 10 2.5 10 0.[2( ) ( 1)( )] 2 5[5 4.5] 4 .5 5 7 S S S = + − = + = MHR • Pre-Calculus 11 Solutions Chapter 1 Page 21 of 80 Section 1.2 Page 27 Question 7 a) t1 = 4, tn = 996, d = 4 Step 1: Determine n. Substitute into tn = t1 + (n − 1)d. 996 = 4 + (n − 1)4 996 = 4 + 4n − 4 996 = 4n n = 249 Step 2: Determine the sum of the series. Substitute into 1( ) 2n n nS t t= + . 4 249 2 9 249 4 996( ) 2 124 500 S S = + = The sum of all the multiples of 4 between 1 and 999 is 124 500. b) t1 = 12, tn = 996, d = 6 Step 1: Determine n. Substitute into tn = t1 + (n − 1)d. 996 = 12 + (n − 1)6 990 = 6n n = 165 Step 2: Determine the sum of the series. Substitute into 1( ) 2n n nS t t= + 6 165 1 5 165 6 99( ) 2 82 665 6S S = + = The sum of all the multiples of 6 between 6 and 999 is 82 665. Section 1.2 Page 28 Question 8 The number of chimes in a 24-h period will be double the sum of the series 1 + 2 + 3 + …+ 12. Substitute t1 = 1, tn = 12, n = 12 into 1( ) 2n n nS t t= + . 1 12 2 12 1( ) 2 12 78 S S = + = In a 24-period the clock will chime 78(2) or 156 times. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 22 of 80 Section 1.2 Page 28 Question 9 a) t5 = 14, d = 3, n = 5 Substitute into tn = t1 + (n − 1)d. 14 = t1 + (5 – 1)3 14 = t1 + 12 t1 = 2 The pilot flew 2 circuits on the first day. b) Total by the end of the fifth day = 2 + 5 + 8 + 11 + 14 = 40 The pilot flew a total of 40 circuits by the end of the fifth day. c) 1[2 ( 1) ] 2 [2( ) ( 1) ] 2 [4 3 3] 2 [1 2 ] 2 3 3 n n n n nS t n d nS n nS n nS n = + − = + − = + − = + The total number of circuits by the end of the nth day is given by (1 3 ) 2 n n+ . Section 1.2 Page 28 Question 10 t2 = 40, t5 = 121 Step 1: Determine the values of t1 and d. Substitute into tn = t1 + (n − 1)d. For t2: 40 = t1 + d For t5: 121 = t1 + 4d Subtracting –81 = –3d 27 = d Then, t1 = 40 − 27 t1 = 13 Step 2: Determine S25. Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 25 25 25 [2( ) ( 1)2 ] 2 25[26 648] 2 8 5 13 25 2 5 7 42 S S S = + − = + = The sum of the first 25 terms of the series is 8425. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 25 of 80 c) Substitute into 1( ) 2n n nS t t= + . 20 2 20 0 ( ) 2 10(35.2) 35 20 6 29 2 .2S S S = + = = Section 1.2 Page 29 Question 16 There are 18 rings of diameter 20 cm down to diameter 3 cm. There are 17 overlaps of 2 cm each. So, find the sum of the 18 diameters and subtract the overlap. Substitute t1 = 20, tn = 3, and n = 18 into 1( ) 2n n nS t t= + . 1 18 8 18 20 207 3( ) 2 S S = + = Then, the distance from the top of the top ring to the bottom of the lowest ring is 207 − 2(17) or 173 cm. Section 1.2 Page 29 Question 17 a) True. If each term is doubled, then 1( ) 2n n nS t t= + will become 1(2 2 ) 2n n nS t t= + which is the same as twice the original sum. b) False. The first sum is 1[2 ( 1) ] 2n nS t n d= + − and double this would be 2Sn = n[2t1 + (n − 1)d] 2Sn = 2nt1 + n2d −nd However, the second sum will be 2 1 2 2 1 2 [2 (2 1) ] 2 2 2 n n nS t n d S nt n d nd = + − = + − Comparing the second sum is n2d greater than the first. c) True. An arithmetic sequence has the form, t1, t1 + d, t1 + 2d, … Multiplying each term by a constant, k, gives kt1, kt1 + kd, kt1 + 2kd, …which has the same pattern just with first term kt1 and common difference kd. For example, 2, 5, 8, 11, … has t1 = 2 and d = 3. When k = 5, the sequence becomes 10, 25, 40, 55, …. Here t1 = 10 and d = 15. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 26 of 80 Section 1.2 Page 30 Question 18 a) Substitute n = 1 into Sn = 2n2 + 5n to determine S1 = 7 = t1 Substitute n = 2, S2 = 18. Then, t2 = 18 – 7 = 11. So d = 4 and t3 = 15. The first three terms of the series are 7 + 11 + 15. b) Substitute in 1[2 ( 1) ] 2n nS t n d= + − . 1 10 0 10 [2( ) ( 1) ] 2 5[14 36] 25 10 7 1 4 0 0S S S = + − = + = c) Substitute in Sn = 2n2 + 5n. S10 = 2(10)2 + 5(10) S10 = 2(100) + 50 S10 = 250 d) 1 2 [2 ( 1) ] 2 [2( ) ( 1) ] 2 [14 4 4] 2 [10 4 7 4 ] 2 5 2 n n n n n nS t n d nS n nS n nS n S n n = + − = + − = + − = + = + Section 1.2 Page 30 Question 19 a) Amount by the end of the 7th hour = 240 + 250 + 260 + 270 + 280 + 290 + 300 b) Substitute into 1[2 ( 1) ] 2n nS t n d= + − . 2 [2( ) ( 1) ] 2 [480 1 2 0 10] 2 [470 10 ] 2 235 4 5 0 10n n n n nS n nS n nS n S n n = + − = + − = + = + MHR • Pre-Calculus 11 Solutions Chapter 1 Page 27 of 80 c) S7 = 235(7) + 5(7)2 S7 = 1645 + 245 S7 = 1890 By the end of the 7th hour Nathan has harvested 1890 bushels. d) No assumptions if conditions were as stated. He would be working non-stop from 11:00 a.m. until 6:00 p.m. and there would be no delays for rain or machinery break- down. Section 1.2 Page 30 Question 20 Substitute S15 = 120, t15 = 43, and n = 15 into 1( ) 2n n nS t t= + . 1 1 1 1 ( ) 2 240 15 15120 645 405 15 2 4 7 3t t t t = + = + − = = − Substitute t15 = 43, t1 = –27, and n = 15 into tn = t1 + (n − 1)d. 43 = –27 + (15 – 1)d 43 = −27 + 14d 70 = 14d d = 5 The first three terms of the series are −27 + (−22) + (−17). Section 1.2 Page 30 Question 21 The formula that Pierre used is, in effect, the same as the one that Jeannette used. In the first formula, substitute tn = t1 + (n − 1)d. 1 1 11 ( ( ) 2 [ ] 2 [2 ( 1 1) ) ] 2 n n n n nS t t nS t nS t n d t n d+ − = + = + = + − MHR • Pre-Calculus 11 Solutions Chapter 1 Page 30 of 80 Section 1.3 Page 39 Question 3 a) Substitute t1 = 2 and r = 3 into tn = t1rn − 1. t2 = 2(3)1 t3 = 2(3)2 t4 = 2(3)3 t2 = 6 t3 = 18 t4 = 54 The first four terms of the sequence are 2, 6, 18, 54. b) Substitute t1 = −3 and r = −4 into tn = t1rn − 1. t2 = −3(−4)1 t3 = −3(−4)2 t4 = −3(−4)3 t2 = 12 t3 = −48 t4 = 192 The first four terms of the sequence are −3, 12, −48, 192. c) Substitute t1 = 4 and r = −3 into tn = t1rn − 1. t2 = 4(−3)1 t3 = 4(−3)2 t4 = 4(−3)3 t2 = −12 t3 = 36 t4 = −108 The first four terms of the sequence are 4, −12, 36, −108. d) Substitute t1 = 2 and r = 0.5 into tn = t1rn − 1. t2 = 2(0.5)1 t3 = 2(0.5)2 t4 = 2(0.5)3 t2 = 1 t3 = 0.5 t4 = 0.25 The first four terms of the sequence are 2, 1, 0.5, 0.25. Section 1.3 Page 39 Question 4 Substitute t1 = 8.1, n = 5, and tn = 240.1 into tn = t1rn − 1. 240.1 = 8.1r4 4 4 240.1 8.1 240.1 8.1 2.333... r r r = = = t2 = 8.1(2.333…)1 t3 = 8.1(2.333…)2 t4 = 8.1(2.333…)3 t2 = 18.9 t3 = 44.1 t4 = 102.9 The missing terms, t2, t3, and t4, are 18.9, 44.1, and 102.9. Section 1.3 Page 39 Question 5 a) Substitute r = 2 and t1 = 3 into tn = t1rn − 1. tn = 3(2)n − 1 b) From the pattern of the sequence, t1= 192 and r = 1 4 − . Substitute into tn = t1rn −1. 11192 4 n nt − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ −= MHR • Pre-Calculus 11 Solutions Chapter 1 Page 31 of 80 c) Substitute into tn = t1rn − 1. For t3: 5 = t1r2 1 2 5t r = For t6: 135 = t1r5 5 2 3 3 5135 135 5 27 3 r r r r r ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ = = = Then, 1 2 5 5 9 t r = = . So, tn = 5 9 (3)n − 1. d) Substitute t1 = 4, n = 13, and tn = 16 384 into tn = t1rn − 1. 16 384 = 4r12 r12 = 4096 12 4096 2 r r = = So, tn = 4(2)n − 1. Section 1.3 Page 39 Question 6 a) Substitute t1 = 5, r = 3, and tn = 135 into tn = t1rn − 1. 135 = 5(3)n − 1 3n − 1 = 27 3n − 1 = 33 So, n − 1 = 3 n = 4 The number of terms, n, is 4. b) Substitute t1 = −2, r = −3, and tn = −1458 into tn = t1rn − 1. −1458 = −2(−3)n − 1 729 = (−3)n − 1 (−3)6 = (−3)n − 1 So, 6 = n − 1 n = 7 The number of terms, n, is 7. c) Substitute t1 = 1 3 , r = 1 2 , and tn = 1 48 into tn = t1rn − 1. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 32 of 80 1 1 4 1 1 1 16 2 1 1 1 1 48 3 1 2 2 So, 4 5 2 1 n n n n n − − − ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = − = The number of terms, n, is 5. d) Substitute t1 = 4, r = 4, and tn = 4096 into tn = t1rn − 1. 4096 = 4(4)n − 1 1024 = (4)n − 1 (4)5 = (4)n − 1 So, 5 = n − 1 n = 6 The number of terms, n, is 6. e) Substitute t1 = 1 6 − , r = 2, and tn = 128 3 − into tn = t1rn − 1. 1 1 1 8 1 ( ) 128 16 6 (2) 3 6 256 2 2 2 128 1 2 3 6 n n n n − − − − = ⎛ ⎞ ⎛ ⎞− − = − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − ⎝ ⎠ = = − So, comparing exponents, n = 9. The number of terms, n, is 9. f) Substitute t1 = 2 2 p , r = 2 p , and tn = 9 256 p into tn = t1rn − 1. 12 17 7 1 9 128 2 2 2 So 256 2 7 8 2 , 1 n n n p p p p n p n p p − − − ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = − = The number of terms, n, is 8. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 35 of 80 Section 1.3 Page 41 Question 12 a) The sequence of terms for the first five days is 1, 2, 4, 8, 16. b) tn = 2n − 1 c) t30 = 229 t30 = 536 870 912 On day 30, Rani would receive 536 870 912 grains of rice. Section 1.3 Page 41 Question 13 a) t1 = 191.41, t2 = 197.34, t3 = 203.46 197.34 191.41 1.0309... r r = = or 203.46 197.34 1.0310... r r = = Each jump Georges improved his performance by a ratio of 1.031, to three decimal places. b) Find t5. Substitute in tn = t1rn − 1. t5 = 191.41(1.031)4 t5 = 216.271… Georges’ winning jump was 216.3 cm, to the nearest tenth. c) Find n when tn = 10 200 1020 = 197.41(1.031)n − 1 1 1 1020 (1.031) 197.41 5.166 911... (1.031) n n − − = = Try values for n: n = 51 (1.031)50 = 4.601… Too small. n = 61 (1.031)60 = 6.244… Too big. n = 56 (1.031)55 = 5.360… Still a bit too big. n = 55 (1.031)54 = 5.199 Too small. If Georges continued to increase his jumps in the same geometric sequence, he would beat Santjie’s record on the 56th jump. Section 1.3 Page 41 Question 14 a) The cell growth of yeast follows the sequence 1, 2, 4, 8, 16, 32. b) tn = 2n − 1 c) Substitute n = 26, for 25 doublings. t26 = 225 MHR • Pre-Calculus 11 Solutions Chapter 1 Page 36 of 80 t26 = 33 554 432 After 25 doublings, there would be 33 554 432 cells. d) The assumption is that all cells continue living. Section 1.3 Page 42 Question 15 t1 = 700, t38 = 2000 37 1 1 37 2000 700 1.0 2000 700 287... n nt t r r r r −= = = = The growth rate was 2.9%, to the nearest tenth of a percent. Section 1.3 Page 42 Question 16 t1 = 2, t2 = 4, t3 = 8 The given terms form a geometric sequence with t1 = 2 and r = 2. So, tn = 2(2)n – 1. Determine n when tn = 142. 142 = 2(2)n − 1 71 = (2)n − 1 Test values of n: Try n = 8 27 = 128 Too big. Try n = 7 26 = 64 Too small. The required number, 71, is between 7 and 8. So, it took Jason 8 weeks to reach his competition number of 142 sledges. Section 1.3 Page 42 Question 17 r = 0.96, t20 = 30 Solve for t1 in tn = t1(0.96)n − 1, when n = 20. 30 = t1(0.96)19 1 19 1 30 (0.96) 65.158... t t = = The arc length for the first swing is 65.2 m, to the nearest tenth of a metre. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 37 of 80 Section 1.3 Page 43 Question 18 t1 = 60, t50 = 1 Substitute and solve for r in tn = t1rn − 1. 1 = 60(r50 −1) 60 = r49 49 1 60 0.9198... r r = = The common ratio for the decrease in doll size is 0.920, to three decimal places. Section 1.3 Page 43 Question 19 a) t1 = 250, r = 100% − 18% or 0.82, 12 or 6 2 n = Since the amount is decreasing every 2 h, we need to use n = 7. Substitute and solve for t7 in tn = t1rn − 1. t7 = 250(0.82)6 t7 = 76.001… After 12 h approximately 76.0 mL of the medicine remains. b) Substitute and solve for n, when tn = 20. 20 = 250(0.82)n − 1 1 1 20 (0.82) 250 0.08 (0.82) n n − − = = Test values for n. Try n = 20 (0.82)19 = 0.023… Too low. Try n = 10 (0.82)9 = 0.167… Too high. Try n = 13 (0.82)12 = 0.0924… Close, but a bit too high. Try n = 14 (0.82)13 = 0.0757… Is now less than the target. The reduction takes place every 2 h, so the amount will be less than 20 mL after approximately 26 h. Section 1.3 Page 43 Question 20 a) Time, d (days) Charge Level, C (%) 0 100 1 100(0.98) = 98 2 100(0.98)2 = 96.04 3 100(0.98)3 = 94.1192 b) tn = 100(0.98)n − 1 MHR • Pre-Calculus 11 Solutions Chapter 1 Page 40 of 80 Section 1.3 Page 44 Question 25 Mala’s solution is correct. Alex’s method is fine, but he did not determine the correct value for r. If the aquarium loses 8% each day, then the next term is 92% of the previous term. This means r = 0.92. Paul’s method is incorrect because the water is evaporating at a constant rate, not a constant amount, each day. The sequence is geometric. Section 1.3 Page 45 Question 26 1 500 50 3 1 100 1 10 1 10 1 20 2 6 18 54 1 16 1 4 1 4 9 5 4 8 3 2 25 4 16 4 1 1 4 1 16 1 64 125 4 32 625 4 100 64 Section 1.3 Page 45 Question 27 a) Area of red shaded portion = area of square − area of circle Area = 2 × 2 – π(1)2 Area = 4 − π Area = 0.8584… The area of the red shaded portion is 0.86 cm2, to the nearest hundredth of a square centimetre. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 41 of 80 b) First find the side length of the blue square. Consider one right triangle that has the side of the red square as its hypotenuse. Let each of its equal legs be x. Use the Pythagorean Theorem. x2 + x2 = 22 2x2 = 4 2x = Then, the side length of the blue square is double x, or 2 2 cm. Area of blue shaded portion = area of larger square − area of larger circle Area = 2 2 × 2 2 − π( 2 )2 Area = 8 − 2π Area = 1.7168… The area of the blue shaded portion is 1.72 cm2, to the nearest hundredth of a square centimetre. c) As in part b), find the side length of the orange square by considering one of the right triangles on its corners. Let y represent each equal leg length. y2 + y2 = ( 2 2 )2 2y2 = 8 y2 = 4 y = 2 Then, area of orange shaded region = area of orange square − area of largest circle Area = 4 × 4 − π(2)2 Area = 3.4336… The area of the orange shaded region is 3.43 cm2, to the nearest hundredth of a square centimetre. d) Explore whether the area is an arithmetic or geometric sequence. t1 = 0.86, t2 = 1.72, t3 = 3.43 This is not an arithmetic sequence, as the difference between successive terms is not constant. Consider the ratio of successive terms: 1.72 2 0.86 = 3.43 1.994 186... 1.72 = Allowing for rounding errors, it appears that each area is 2 times the previous area. In tn = t1rn − 1, substitute r = 2, t1 = 4 − π and n = 8. t8 = (4 − π)(2)7 t8 = 109.876… The area of the newly shaded part of the 8th square would be 109.88 cm2, to the nearest hundredth of a square centimetre. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 42 of 80 Section 1.4 Geometric Series Section 1.4 Page 53 Question 1 a) The series 4 + 24 + 144 + 864 + … is geometric because each term is 6 times the previous term. b) The series −40 + 20 − 10 + 5 − … is geometric because each term is −0.5 times the previous term. c) The series 3 + 9 + 18 + 54 + … is not geometric. d) The series 10 + 11 + 12.1 + 13.31 + … is geometric because each term is 1.1 times the previous term. Section 1.4 Page 53 Question 2 a) t1 = 6, r = 3 2 , n = 10 Substitute into 1( 1) 1 n n t rS r − = − to find S10. ( ) 10 1 10 10 0 10 1 3 1 2 59 0496 1 3 1024 1 2 58 0256 2 1024 174 07 2 5 256 6 S S S S ⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦= − ⎡ ⎤−⎢ ⎥⎣ ⎦= ⎡ ⎤= ⎢ ⎥⎣ ⎦ = S10 = 679.98, to the nearest hundredth. b) t1 = 18, r = 1 2 − , n = 12 Substitute into 1( 1) 1 n n t rS r − = − to find S12. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 45 of 80 d) Substitute t1 = 72, r = 1 2 , n = 12 into 1( 1) 1 n n t rS r − = − . 12 12 1 12 2 1 1 172 1 4096 172 2 1 2 1 2 72 S S S ⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦= − ⎛ ⎞⎡ ⎤−⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠= − = 9 4095 4096 − 512 256 2 − ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ( ) 12 36 855 256 S = Section 1.4 Page 53 Question 4 a) Substitute t1 = 27, tn = 1 243 , r = 1 3 into 1 1 n n rt tS r − = − . 1 1 327 729 2 19 682 3 729 2 9 1 1 841 or 40.50 243 27 3 243 1 3 n n n n n S S S S S ⎛ ⎞⎛ ⎞ −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= − ⎡ ⎤⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = ≈ MHR • Pre-Calculus 11 Solutions Chapter 1 Page 46 of 80 b) Substitute t1 = 1 3 , tn = 128 6561 , r = 2 3 into 1 1 n n rt tS r − = − . ( ) 1 6305 3 19 6 2 128 1 3 6561 3 2 83 6305 or 0.96 65 1 3 6 n n n n S S S S ⎛ ⎞⎛ ⎞ −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= − −⎛ ⎞= −⎜ ⎟ ⎝ ⎠ = ≈ c) Substitute t1 = 5, tn = 81 920, r = 4 into 1 1 n n rt tS r − = − . 4 81 920 5( ) 1 109 2 4 25 n n S S − = − = d) Substitute t1 = 3, tn = 46 875, r = –5 into 1 1 n n rt tS r − = − . 5 46 875 3( ) 15 39 063 n n S S = − −− − = Section 1.4 Page 54 Question 5 a) Substitute Sn = 33, tn = 48, r = −2 into 1 1 n n rt tS r − = − . 1 1 1 1 ( ) 1 3(33 2 483 ) 96 99 9 3 6 3 2 t t t t − − = − = + − = = − MHR • Pre-Calculus 11 Solutions Chapter 1 Page 47 of 80 b) Substitute Sn = 443, n = 6, r = 1 3 into 1( 1) 1 n n t rS r − = − . 1 6 1 1 1 1 1 1 1 1 3443 1 729 2 728 3443 729 2 364443 243 243 1 3 443 1 3 443 364 295.7 t t t t t t ⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦= − ⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ −⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ≈ Section 1.4 Page 54 Question 6 Substitute Sn = 4372, t1 = 4, r = 3 into 1( 1) 1 n n t rS r − = − . 7 ( 1) 1 4372 2(3 1) 2186 3 1 4 34372 3 2187 3 3 3 7 n n n n n n − = − = − = − = = = There are seven terms in the series. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 50 of 80 5 5 5 16 0.4 0 [( ) 1] 1 26.393 4 6 . S S − = − = The total distance travelled when the ball hits the ground for the sixth time is 20 m + 26.39 m or 46.4 m, to the nearest tenth of a metre. Section 1.4 Page 54 Question 11 Substitute t1 = 25, r = 1.1, n = 15 into 1( 1) 1 n n t rS r − = − . 1 15 1 5 5 25 1.1 1. [( ) 1] 1 794.312... 1 S S − = − = By the end of the 15th week, Celia will have run 794.3 km, to the nearest tenth of a kilometre. Section 1.4 Page 54 Question 12 a) b) Stage Number Length of Each Line Segment Number of Line Segments Perimeter of Snowflake 1 1 3 3 2 1 3 12 4 3 1 9 48 16 3 4 1 27 192 64 9 5 1 81 768 256 27 MHR • Pre-Calculus 11 Solutions Chapter 1 Page 51 of 80 c) In each of the following general terms, n is the stage number. Length of each line segment: tn = 11 3 n− ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Number of line segments: tn = 3(4)n − 1 Perimeter of snowflake = 143 3 n− ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ d) When n = 6, perimeter 54 10243 or 3 81 ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ . The total perimeter at stage 6 is 1024 81 , or approximately 12.64. Section 1.4 Page 55 Question 13 Substitute t1 = 1000(1.4), r = 1.4, and n = 100 10 or 10 into 1( 1) 1 n n t rS r − = − . 1 10 0 10[( ) 1] 1 1000(1.4) 1.4 1.4 97 739.129... S S − = − = After 100 days, 98 739 people will be aware of the product. Section 1.4 Page 55 Question 14 Substitute t1 = 24, r = 3 4 or 0.75, and n = 10 into 1( 1) 1 n n t rS r − = − . 1 10 10 0 24 0.75 0 [( ) 1] 1 90.593... .75 S S − = − = The total length of the line of 10 beads was 91 mm, to the nearest millimetre. Section 1.4 Page 56 Question 15 a) t1 = 200, t2 = 200 + 200(0.12) t3 = 200 + [200 + 200(0.12)](0.12) t3 = 200 + 200(0.12) + 200(0.12)2 t3 = 226.88 The amount of ampicillin in the body after taking the third tablet is 226.9 mg, to the nearest tenth. b) Substitute t1 = 200, r = 0.12, n = 6 into 1( 1) 1 n n t rS r − = − . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 52 of 80 6 6 6 200 0.12 0. [( ) 1] 1 227.272... 12 S S − = − = The amount of ampicillin in the body after taking the sixth tablet is 227.3 mg, to the nearest tenth. Section 1.4 Page 56 Question 16 Substitute t1 = 3, r = 3, and Sn = 9840 into 1( 1) 1 n n t rS r − = − . 8 ( 1) 1 2(9840) 3 1 3 6560 1 3 6561 3 3 3 8 3 39840 3 n n n n n n − = − = − + = = = = The series has eight terms. Section 1.4 Page 56 Question 17 t3 = t1(r2) = 24 and t4 = t1(r3) = 36 So, r = 36 24 or 3 2 . Then, substituting in t3: 2 1 1 1 24 24(4) 9 32 3 3 2 t t t ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ = = Substitute t1 = 32 3 , r = 3 2 , and n = 10 into 1( 1) 1 n n t rS r − = − . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 55 of 80 Section 1.4 Page 57 Question 22 Answers may vary. Examples: a) Tom is assuming that all 400 eggs do in fact produce a butterfly. b) Tom’s assumption is very optimistic. Some eggs would not survive due to weather, predators, or other unfavourable circumstances. Also he has calculated the total number in the first to fifth generations, not the fifth generation only. c) Tom’s estimate is probably way too much. See the reasons given in part b). d) I would research the life span of the butterflies and the normal number of successful hatchings expected from 400 eggs. Section 1.5 Infinite Geometric Series Section 1.5 Page 63 Question 1 a) Since r > 1, the series is divergent. b) Since −1 < r < 1, the series is convergent. c) 1 5 r = ; since −1 < r < 1, the series is convergent. d) r = 2; since r >1, the series is divergent. e) 9 27 5 25 9 25 5 27 5 3 r r r ⎛ ⎞ ⎛ ⎞= − ÷⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = − Since r < −1, the series is divergent. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 56 of 80 Section 1.5 Page 63 Question 2 a) Substitute t1 = 8 and r = 1 4 − into 1 1 tS r∞ = − . 1 1 1 48 5 32 2 or 6 5 1 4 5 8 tS r S S S ∞ ∞ ∞ ∞ = − = ⎛ ⎞− ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ = b) Since 4 3 r = , r > 1 and the series is divergent and has no infinite sum. c) If r = 1, the series is not geometric. The series would just be 5 + 5 + 5 + …. It has no infinite sum. d) 1 2 r = . Substitute t1 = 1 and r = 1 2 into 1 1 tS r∞ = − . 1 1 2 1 2 S S ∞ ∞ = − = e) 12 4 5 12 1 5 4 3 5 r r r = − ÷ ⎛ ⎞= − ⎜ ⎟ ⎝ ⎠ = − Substitute t1 = 4 and r = 3 5 − into 1 1 tS r∞ = − . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 57 of 80 8 1 54 8 5 1 or 2 4 5 2 2 3 S S S ∞ ∞ = ⎛ ⎞− ⎜ ⎟ ⎝ ⎠ ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ = − Section 1.5 Page 63 Question 3 a) 0.87 0.878 787 87... 0.87 0.0087 0.000 087 ... = = + + + Substitute t1 = 87 100 and r = 1 100 into 1 1 tS r∞ = − . 1 87 100 100 99 87 29 or 99 33 87 100 1 100 S S S ∞ ∞ ∞ = − ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = b) 0.437 0.437 437 437... 0.437 0.000 437 0.000 000 437 ... = = + + + Substitute t1 = 437 1000 and r = 1 1000 into 1 1 tS r∞ = − . 1 437 4 1000 1000 37 100 999 0 1 1 437 99 000 9 S S S ∞ ∞ ∞ = − ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = MHR • Pre-Calculus 11 Solutions Chapter 1 Page 60 of 80 b) The assumption is that the trend does continue and that the well is kept operational until it runs dry. This is not reasonable: once production is low, the well would not be profitable to operate and would probably be closed. Section 1.5 Page 64 Question 9 Substitute t1 = 1, r = 3x, and S∞ = 4 into 1 1 tS r∞ = − . 1 14(1 3 ) 1, 3 3 12 1 1 3 4 4 x x x x x = − − = ≠ = = The value of x is 1 4 . The first four terms of the series are 3 9 271 4 16 64 + + + . Section 1.5 Page 64 Question 10 Let t1 = x, then S∞ = 2x. Substitute into 1 1 tS r∞ = − . 1 1 2 11 2 1 2 2 r xr x x r r x = − − = − = = Section 1.5 Page 64 Question 11 A series is convergent if −1 < r < 1. a) r = x, so the series is convergent if −1 < x < 1. b) 3 xr = , so the series is convergent if 1 1 or 3 3. 3 x x− < < − < < c) r = 2x, so the series is convergent if −1 < 2x < 1 or 1 1 2 2 x− < < . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 61 of 80 Section 1.5 Page 64 Question 12 The perimeter of the largest triangle is 3 cm. Each smaller triangle is half this. Substitute t1 = 3 and r = 1 2 into 1 1 tS r∞ = − . 3 1 2 1 6 S S ∞ ∞ = − = The sum of the perimeters in this infinite series is 6 cm. Section 1.5 Page 64 Question 13 Substitute t1 = 50 and r = 0.8 into 1 1 tS r∞ = − . 50 0.1 250 8 S S ∞ ∞ = − = The pendulum will swing a total of 250 cm. Section 1.5 Page 64 Question 14 Andrew’s answer is not reasonable. The series is not convergent as r = 1.1, which is greater than 1. The series has no sum. Just looking at the terms, each term is greater than 1 so the sum of the first five or six terms is already greater than 10. Section 1.5 Page 64 Question 15 The pattern of vertical heights is: 16, 2(16) 1 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , 2(16) 21 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , 2(16) 31 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , 2(16) 41 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , … One way to find the sum of all vertical heights is to find the sum of the downward distances and the sum of the upward distances. The downward sum has one extra term, 16. For the downward sum, substitute t1 = 16 and r = 1 2 into 1 1 tS r∞ = − . 16 1 2 1 32 S s ∞ ∞ = − = For the upward sum, substitute t1 = 8 and r = 1 2 into 1 1 tS r∞ = − . MHR • Pre-Calculus 11 Solutions Chapter 1 Page 62 of 80 8 1 2 1 16 S s ∞ ∞ = − = The total vertical distance that the ball travels is 32 + 16, or 48 m. Section 1.5 Page 64 Question 16 a) If the sequence is geometric, then 27 9or 30 10 r = . Substitute into 1( 1) 1 n n t rS r − = − to find S8. 8 8 8 1 1 170.859. 930 . 10 9 1 . 0 S S ⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦= − = After 8 times, the post is pounded 170.86 cm into the ground, to the nearest hundredth of a centimetre. b) Substitute t1 = 30 and r = 9 10 into 1 1 tS r∞ = − . 30 9 1 1 300 0 S S ∞ ∞ = − = If the post is pounded indefinitely, it will be pounded 300 cm into the ground. Section 1.5 Page 64 Question 17 a) Rita is correct. b) r = 4 1 or 1 3 3 − − . Since r < −1 the series is divergent and has no sum. Section 1.5 Page 64 Question 18 Substitute t1 = 25 and r = 0.8 into 1 1 tS r∞ = − . 25 0.1 125 8 S S ∞ ∞ = − = The balloon’s maximum altitude would be 125 m. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 65 of 80 c) For the sum of an infinite geometric series substitute into 1 1 tS r∞ = − . 1 1 4 1 4 3 S S ∞ ∞ = − = Section 1.5 Page 65 Question 23 Step 3 n 1 2 3 4 Fraction of Paper 1 4 1 16 1 64 1 256 Step 4 Area received by each student is 1 1 1 1 4 16 64 256 + + + . The areas form an infinite geometric series, with t1 = 1 4 and r = 1 4 . 1 1 4 4 1 4 1 4 3 1 3 S S S ∞ ∞ ∞ = − ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = Chapter 1 Review Chapter 1 Review Page 66 Question 1 a) arithmetic, d = 4 b) arithmetic, d = −5 c) not arithmetic d) not arithmetic MHR • Pre-Calculus 11 Solutions Chapter 1 Page 66 of 80 Chapter 1 Review Page 66 Question 2 a) From the pattern of the sequence, t1 = 18 and d = 12. Substitute into tn = t1 + (n − 1)d. tn = 18 + (n − 1)12 tn = 12n + 6 This matches C. b) From the pattern of the sequence, t1 = 7 and d = 5. Substitute into tn = t1 + (n − 1)d. tn = 7 + (n − 1)5 tn = 5n + 2 This matches D. c) From the pattern of the sequence, t1 = 2 and d = 2. Substitute into tn = t1 + (n − 1)d. tn = 2 + (n − 1)2 tn = 2n This matches E. d) From the pattern of the sequence, t1 = −8 and d = −4. Substitute into tn = t1 + (n − 1)d. tn = −8 + (n − 1)(−4) tn = −4n −4 tn = −4(n + 1) This matches B. e) From the pattern of the sequence, t1 = 4 and d = 3. Substitute into tn = t1 + (n − 1)d. tn = 4 + (n − 1)3 tn = 3n + 1 This matches A. Chapter 1 Review Page 66 Question 3 t1 = 7, d = 7 a) All of the terms are multiples of 7 and 98 is a multiple of 7. Substitute into tn = t1 + (n − 1)d and solve for n. 98 = 7 + (n − 1)7 98 = 7n n = 98 7 n = 14 b) Since 110 is not a multiple of 7, it is not a term of this sequence. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 67 of 80 c) 378 is a multiple of 7. n = 378 7 n = 54 d) Since 575 is not a multiple of 7, it is not a term of this sequence. Chapter 1 Review Page 66 Question 4 a) Both sequences are arithmetic. For sequence 1: t1 = 2 and d = 7. Substitute n = 17 into tn = t1 + (n − 1)d to determine t17. t17 = 2 + (17 − 1)7 t17 = 114 For sequence 2: t1 = 4 and d = 6. Substitute n = 17 into tn = t1 + (n − 1)d to determine t17. t17 = 4 + (17 − 1)6 t17 = 100 Statement A is correct, t17 is greater in sequence 1. b) Yes, the graph supports the answer in part a). The graph shows that sequence 1 is growing at a faster rate than sequence 2. The two sequences have the same third term, 16, then further terms are greater in sequence 1. Chapter 1 Review Page 66 Question 5 t1 = 5, t4 = 17 Substitute for t4 in tn = t1 + (n − 1)d and solve for d. 17 = 5 + (4 − 1)d 12 = 3d d = 4 Then, determine t10. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 70 of 80 c) Substitute Sn = 625, t1 = 1, and d = 2 into 1[2 ( 1) ] 2n nS t n d= + − and solve for n. 625 1[2( ) ( )21 ] 2 n n= + − 625 = n2 n = 25 You would need 25 days to contact all 625 people in the neighbourhood. Chapter 1 Review Page 67 Question 10 a) The sequence of number of squares is 5, 5 + 4, 5 + 2(4), 5 + 3(4), … The total number of squares in the 15th step will be: 5 + 14(4) = 61 There are 61 squares in the 15th step of the design. b) t1 = 5, tn = 61, n = 15 Substitute into 1( ) 2n n nS t t= + . 1 15 5 15 5 6( ) 2 4 1 95 S S = + = To build all 15 steps, 495 squares are needed. Chapter 1 Review Page 67 Question 11 Substitute t1 = 10, d = 2 and n = 30 into 1[2 ( 1) ] 2n nS t n d= + − . 30 30 10[2( ) (30 1) ] 2 2S = + − S30 = 15[20 + 29(2)] S30 = 1170 The entire concert hall has 1170 seats. Chapter 1 Review Page 67 Question 12 a) Not geometric, because there is no common ratio between successive terms. b) geometric: t1 = 1, r = −2, tn = (−2)n − 1 MHR • Pre-Calculus 11 Solutions Chapter 1 Page 71 of 80 c) geometric: t1 = 1, r = 1 2 , tn = 11 2 n− ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ d) Not geometric, because there is no common ratio between successive terms. Chapter 1 Review Page 67 Question 13 a) t1 = 5000, r = 100% + 8% or 1.08 At the end of the 5th hour, the 6th term will occur. Substitute for t6 in tn = t1rn −1. t6 = 5000(1.08)5 t6 = 7346.64… The number of bacteria present at the end of 5 h is 7346. b) tn = t1rn −1 tn = 5000(1.08)n − 1 A formula for the number of bacteria present at the start of the nth hour is 5000(1.08)n − 1. So after the nth hour, the number is 5000(1.08)n − 1(1.08) or 5000(1.08)n. Chapter 1 Review Page 67 Question 14 Stage 1: radius 181 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or 27 cm Stage 2: radius 2181 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or 9 cm Stage 3: radius 3181 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or 3 cm Stage 4: radius 4181 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or 1 cm So, C = 2π(1) = 2π or approximately 6.28. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 72 of 80 Chapter 1 Review Page 67 Question 15 Answers will vary. Arithmetic Sequence Geometric Sequence ↓ ↓ Definition: A pattern of numbers in which there is a constant difference between successive terms Definition: A pattern of numbers in which there is a constant ratio between successive terms ↓ ↓ Formula: tn = t1 + (n − 1)d Formula: tn = t1rn − 1 ↓ ↓ Example: 4, 7, 10, 13, … tn = 4 + (n − 1)3 tn = 1 + 3n Example: 4, 8, 16, 32, … tn = 4(2)n − 1 Chapter 1 Review Page 67 Question 16 a) arithmetic series b) geometric series c) geometric series d) arithmetic series e) arithmetic series with d = 1 4 f) geometric series with r = 2 3 Chapter 1 Review Page 68 Question 17 a) Substitute t1 = 6, r = 1.5, and n = 10 into 1( 1) 1 n n t rS r − = − . 1 10 0 0 1 6 1.5 1.5 ( 1) 1 679.980... S S − = − = S10 is approximately 679.98. b) Substitute t1 = 18, r = 0.5, and n = 12 into 1( 1) 1 n n t rS r − = − . 1 12 1 2 2 18 0.5 0. ( 1) 1 35.991... 5 S S − = − = S12 is approximately 35.99. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 75 of 80 d) Convergent; substitute t1 = 3 4 and r = 1 2 into 1 1 tS r∞ = − . 1 3 2 4 3 4 1 2 1 3 2 S S S ∞ ∞ ∞ = − ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = Chapter 1 Review Page 68 Question 21 a) r = 2.8 2 or 0.4 7 5 − = − − b) S1 = 7 S2 = 7 − 2.8 = 4.2 S3 = 7 − 2.8 + 1.12 = 5.32 S4 = 7 − 2.8 + 1.12 − 0.448 = 4.872 S5 = 7 − 2.8 + 1.12 − 0.448 + (0.448)(0.4) = 5.0512 c) The sums seem to be approaching 5. d) Substitute t1 = 7 and r = −0.4 into 1 1 tS r∞ = − . 7 0.1 ( ) 5 4 S S ∞ ∞ = −− = The sum of the series is 5. MHR • Pre-Calculus 11 Solutions Chapter 1 Page 76 of 80 Chapter 1 Review Page 68 Question 22 a) Areas of the sequence of squares are: 1, 21 1 2 4 ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ , 21 1 4 16 ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ , 21 1 8 64 ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ The areas form a geometric sequence with t1 = 1 and r = 1 4 . b) Sum of the areas of the four squares is 1 1 1 64 16 4 1 851 or 4 16 64 64 64 + + + + + + = square units. c) Substitute t1 = 1 and r = 1 4 into 1 1 tS r∞ = − . 1 1 4 1 4 3 S S ∞ ∞ = − = The infinite sum of the areas of the squares is 4 1or 1 3 3 square units. Chapter 1 Review Page 68 Question 23 a) • A series is geometric if there is a common ratio r such that r is any real number, r ≠ 1. • An infinite geometric series converges if –1 < r < 1. • An infinite geometric series diverges if r < –1 or r > 1. b) Answers will vary. Examples: Positive common ratio: 3 + 1.5 + 0.75 +…. Substitute t1 = 3 and r = 1 2 into 1 1 tS r∞ = − . 1 8 1 4 1 2 1 MHR • Pre-Calculus 11 Solutions Chapter 1 Page 77 of 80 3 1 2 1 6 S S ∞ ∞ = − = Negative common ratio: 10 − 2 + 2 2 ... 5 25 − + Substitute t1 = 10, r = 1 5 − into 1 1 tS r∞ = − . 1 25 1 or 8 3 10 1 5 3 S S ∞ ∞ = ⎛ ⎞− ⎜ ⎟ ⎝ ⎠ = − Chapter 1 Practice Test Page 69 Question 1 The difference between the terms 3 and 9 is 6. Use d = 6 to determine the other terms. 3 − 6 = −3, 9 + 6 = 15, 15 + 6 = 21 The best answer is D. Chapter 1 Practice Test Page 69 Question 2 The pattern is an arithmetic sequence with t1 = 1 and d = 3. Substitute into tn = t1 + (n − 1)d. tn = 1 + (n − 1)3 tn = 3n − 2 The best answer is B. Chapter 1 Practice Test Page 69 Question 3 First, determine r: r = 343 1 2401 7 = − − . Substitute t1 = 16 807, r = 1 7 − , and n = 5 into 1( 1) 1 n n t rS r − = − . 5 5 5 116 807 7 1 1 14 707 1 7 S S ⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ − = − ⎦ − = The best answer is B.