CHAPTER 19 NUMBER SEQUENCES

Transcription

1 EXERCISE 77 Page 167 CHAPTER 19 NUMBER SEQUENCES 1. Determie the ext two terms i the series: 5, 9, 13, 17, It is oticed that the sequece 5, 9, 13, 17,... progressively icreases by 4, thus the ext two terms will be 1 ad 5. Determie the ext two terms i the series: 3, 6, 1, 4, It is oticed that the sequece 3, 6, 1, 4, progressively doubles, thus the ext two terms will be 48 ad Determie the ext two terms i the series: 11, 56, 8, It is oticed that the sequece 11, 56, 8, progressively halves, thus the ext two terms will be 14 ad 7 4. Determie the ext two terms i the series: 1, 7,, It is oticed that the sequece 1, 7,, progressively decreases by 5, thus the ext two terms will be 3 ad 8 5. Determie the ext two terms i the series:, 5, 10, 17, 6, 37, It is oticed that the sequece, 5, 10, 17, 6, 37, progressively icreases by 3, the 5, the 7, the 9, ad so o. Thus the ext two terms will be = 50 ad = Determie the ext two terms i the series: 1, 0.1, 0.01, It is oticed that the sequece 1, 0.1, 0.01, progressively decreases to oe teth of its previous value, thus the ext two terms will be ad , Joh Bird

2 7. Determie the ext two terms i the series: 4, 9, 19, 34, It is oticed that the sequece 4, 9, 19, 34, progressively icreases by 5, the 10, the 15, ad so o. Thus the ext two terms will be = 54 ad = , Joh Bird

3 EXERCISE 78 Page The th term of a sequece is give by 1. Write dow the first four terms. Whe = 1, 1 = 1 Whe =, 1 = 3 Whe = 3, 1 = 5 Whe = 4, 1 = 7 Hece, the first four terms are 1, 3, 5, 7,.... The th term of a sequece is give by Write dow the first five terms. Whe = 1, = 7 Whe =, = 10 Whe = 3, = 13 Whe = 4, = 16 Whe = 5, = 19 Hece, the first five terms are 7, 10, 13, 16, Write dow the first four terms of the sequece give by The first four terms of the series will be: i.e. 6, 11, 16 ad 1 5(1) + 1, 5() + 1, 5(3) + 1 ad 5(4) Fid the th term i the series: 5, 10, 15, 0, We otice that the gap betwee each of the give four terms is 5, hece the law relatig the umbers is: 5 + somethig The secod term, so whe =, the 10 = 5 + somethig, 10 = 10 + somethig, so the somethig must be 0 Thus, the th term of the series 5, 10, 15, 0, is: 5 5. Fid the th term i the series: 4, 10, 16,, , Joh Bird

4 We otice that the gap betwee each of the give four terms is 6, hece the law relatig the umbers is: 6 + somethig The secod term, so whe =, the 10 = 6 + somethig, 10 = 1 + somethig, so the somethig must be (from simple equatios) Thus, the th term of the series 4, 10, 16,, is: 6 6. Fid the th term i the series: 3, 5, 7, 9,... We otice that the gap betwee each of the give four terms is, hece the law relatig the umbers is: + somethig The secod term, so whe =, the 5 = + somethig, 5 = 4 + somethig, so the somethig must be 1 (from simple equatios) Thus, the th term of the series 3, 5, 7, 9,... is: Fid the th term i the series:, 6, 10, 14,... We otice that the gap betwee each of the give four terms is 4, hece the law relatig the umbers is: 4 + somethig The secod term, so whe =, the 6 = 4 + somethig, 6 = 8 + somethig, so the somethig must be (from simple equatios) Thus, the th term of the series, 6, 10, 14,... is: 4 8. Fid the th term i the series: 9, 1, 15, 18,... We otice that the gap betwee each of the give four terms is 3, hece the law relatig the umbers is: 3 + somethig , Joh Bird

5 The secod term, so whe =, the 1 = 3 + somethig, 1 = 6 + somethig, so the somethig must be 6 (from simple equatios) Thus, the th term of the series 9, 1, 15, 18,... is: Write dow the ext two terms of the series: 1, 8, 7, 64, 15, This is a special series ad does ot follow the patter of the previous examples. Each of the terms i the give series are cubic umbers, i.e. 1, 8, 7, 64, 15, 1 3, 3, 3 3, 4 3, 5 3, Hece the th term is: 3 Thus, the 6th term is: 6 3 = 16 ad the 7th term is: 7 3 = , Joh Bird

6 EXERCISE 79 Page Fid the 11th term of the series 8, 14, 0, 6,... The 11th term of the series 8, 14, 0, 6, is give by: a + ( 1)d where a = 8, = 11 ad d = 6 Hece, the 11th term is: 8 + (11 1)(6) = = 68. Fid the 17th term of the series 11, 10.7, 10.4, 10.1,... The 17th term of the series 11, 10.7, 10.4, 10.1,... is give by: a + ( 1)d where a = 11, = 17 ad d = 0.3 Hece, the 17th term is: 11 + (17 1)( 0.3) = = The seveth term of a series is 9 ad the eleveth term is 54. Determie the sixteeth term. The th term of a arithmetic progressio is: a + ( 1)d The 7th term is: a + 6d = 9 (1) The 11th term is: a + 10d = 54 () () (1) gives: 4d = 5 from which, d = 5 4 Substitutig i (1) gives: a = 9 i.e. a = 9 from which, a = = Hece, the 16th term is: (16 1) 4 = = Fid the 15th term of a arithmetic progressio of which the first term is.5 ad the teth term is 16 The th term of a arithmetic progressio is: a + ( 1)d ad if a =.5 ad the 10th term is 16, the:.5 + 9d = , Joh Bird

7 from which, 9d = 16.5 = 13.5 ad d = 13.5/9 = 1.5 Hece, the 15th term is: a + ( 1)d =.5 + (15 1)(1.5) =.5 + (14)(1.5) = = Determie the umber of the term which is 9 i the series 7, 9., 11.4, 13.6,... 9 = 7 + ( 1)(.) from which, 9 7 =.( 1) i.e.. = 1 i.e. 10 = 1 ad = 11 i.e. 9 is the 11th term of the series 6. Fid the sum of the first 11 terms of the series 4, 7, 10, 13,... I the series 4, 7, 10, 13,... a = 4 ad d = 3 Sum of series, S = [ a+ ( 1) d] 11 S = (4) + (11 1)(3) = = 09 Whe = 11, 11 [ ] [ ] 7. Determie the sum of the series 6.5, 8.0, 9.5, 11.0,, 3 I the series: 6.5, 8.0, 9.5, 11.0,, 3, a = 6.5 ad d = 1.5 The th term is 3, hece, a + ( 1)d = 3 i.e ( 1)(1.5) = 3 from which, = ( 1)(1.5) ad = 1 i.e. 17 = 1 ad = = + = + = + = Sum of series, S [ a ( 1) d] [ (6.5) (18 1)(1.5) ] 9[ ] , Joh Bird

8 EXERCISE 80 Page The sum of 15 terms of a arithmetic progressio is 0.5 ad the commo differece is. Fid the first term of the series. = 15, d = ad S 15 = 0.5. Sice the sum of terms of a AP is give by: S = [a + ( 1)d] The 0.5 = 15 [a + (15 1)] = 15 [a + 8] Hece, = a + 8 i.e. 7 = a + 8 Thus, a = 7 8 = 1 from which, a = i.e. the first term, a = = 0.5. Three umbers are i arithmetic progressio. Their sum is 9 ad their product is 0.5. Determie the three umbers. Let the three umbers be (a d), a ad (a + d) Thus, (a d) + a + (a + d) = 9 i.e. 3a = 9 ad a = 3 Also, a(a d)(a + d) = 0.5 Sice, a = 3, the 3( 9 d ) = i.e. 9 d = = ad = d from which, d =.5 ad d =.5 = 1.5 Hece, the three umbers are: (a d) = = 1.5, a = 3 ad (a + d) = = , Joh Bird

9 3. Fid the sum of all the umbers betwee 5 ad 50 which are exactly divisible by 4 The series 8, 1, 16, 0,, 48 is a AP whose first term a = 8 ad commo differece d = 4 The last term is: a + ( 1)d = 48 i.e. 8 + ( 1)4 = 48 Hece, ( 1) = = = 60 from which, = The sum of all 61 terms is give by: S61 = [ a+ ( 1) d] = [ (8) + (61 1)4] = [ ] = 30.5( 56) = Fid the umber of terms of the series 5, 8, 11,... of which the sum is 105 Sum of terms is give by: S = [ a+ ( 1) d] i.e. 105 = [ (5) + ( 1)(3) ] i.e. 105 = [ ( 1) ] Hece, 050 = [ ] [ ] i.e = = = 7+ 3 This is a quadratic equatio, hece = 7 7 4(3)( 050) ± ± ± = = (3) 6 6 i.e. umber of terms, = 5 (the egative aswer havig o meaig) 5. Isert four terms betwee 5 ad.5 to form a arithmetic progressio. The th term of a arithmetic progressio is: a + ( 1)d ad if a = 5 ad the 6th term is.5, the: 5 + (6 1)d =.5 i.e. 5d =.5 5 = 17.5 from which, the commo differece, d = 17.5/5 = , Joh Bird

10 Hece, the series is: 5, 8.5, 1, 15.5, 19,,5 6. The first, teth ad last terms of a arithmetic progressio are 9, 40.5 ad 45.5, respectively. Fid (a) the umber of terms, (b) the sum of all the terms, ad (c) the 70th term. (a) a = 9 ad the 10th term is: a + (10 1)d = 40.5 i.e d = 40.5 ad 9d = = 31.5 hece 31.5 d = = Last term is give by: a + ( 1)d i.e. 9 + ( 1)(3.5) = 45.5 i.e. ( 1)(3.5) = = ad 1 = = Hece, the umber of terms, = (b) Sum of all the terms, S = [ a+ ( 1) d] = [ (9) + (10 1)(3.5) ] = 60[ ] (c) The 70th term is: a + ( 1)d = 9 + (70 1)(3.5) = (3.5) = 50.5 = O commecig employmet a ma is paid a salary of per aum ad receives aual icremets of 480. Determie his salary i the 9th year ad calculate the total he will have received i the first 1 years. The series is: , , , to 9 terms, i.e. a = , d = 480 ad = 9 Salary after 9 years = a + ( 1)d = (9 1)(480) = = Thus, total salary i 1 years, 1 S = a+ d = + = + [ ( 1) ] [ (16 000) (1 1)(480) ] 6[ ] 30 = 6(37 80) = , Joh Bird

11 8. A oil compay bores a hole 80 m deep. Estimate the cost of borig if the cost is 30 for drillig the first metre with a icrease i cost of per metre for each succeedig metre. The series is: 30, 3, 34, to 80 terms, i.e. a = 30, d = ad = = + ( 1) = (30) + (80 1)() = = 40(18) = 870 Thus, total cost, S [ a d] [ ] [ ] , Joh Bird

12 EXERCISE 81 Page Fid the 10th term of the series 5, 10, 0, 40,... The 10th term of the series 5, 10, 0, 40, is give by: ar 1 where a = 5, r = ad = 10 i.e. the 10th term = 1 5 = (5)() = 5(51) = 560 ar = ( )( ) Determie the sum of the first 7 terms of the series 0.5, 0.75,.5, 6.75, , 0.75,.5, 6.75, is a GP with a commo ratio r = 3 The sum of terms, S a r = ( 1) ( r 1) Hece, the sum of the first 7 terms, S 7 7 ( ) ( ) ( ) ( ) = = = 3 1 = The 1st term of a geometric progressio is 4 ad the 6th term is 18. Determie the 8th ad 11th terms. 18 The 6th term is give by: ar 5 = 18 i.e. 4r 5 = 18 ad r 5 = = 3 4 Thus, r = 5 3 = Hece, the 8th term is: ar 1 = 4() 8 1= 4() 7= 4(18) = 51 ad the 11th term is: 4() 11 1 = 4() 10 = 4(104) = Fid the sum of the first 7 terms of the series, 5, 1.5,... (correct to 4 sigificat figures). ar 5 Commo ratio, r = a = =.5 (also, ar 1.5 ar = 5 =.5) Sum of 7 terms, S ( ) ( r ) 7 ( ) ( ) ( ) a r = = = = 81.5, correct to 4 sigificat figures , Joh Bird

13 5. Determie the sum to ifiity of the series 4,, 1,... The series is a GP where r = 4 = 0.5 ad a = 4 Hece, sum to ifiity, a 4 4 S = 1 r = 1 (0.5) = 0.5 = 8 6. Fid the sum to ifiity of the series 1, 1 1 4, 5 8,... The series is a GP where r = Hece, sum to ifiity, 1.5 = 0.5 ad a =.5.5 a S = = = = = 1 1 r 1 ( 0.5) , Joh Bird

14 EXERCISE 8 Page I a geometric progressio the 5th term is 9 times the 3rd term ad the sum of the 6th ad 7th terms is Determie (a) the commo ratio, (b) the first term ad (c) the sum of the 4th to 10th terms iclusive. (a) The 5th term of a geometric progressio is: ar 4 ad the 3rd term is: ar r 4 Hece, ar 4 = 9 ar from which, = 9 i.e. r = 9 r from which, the commo ratio, r = 3 (b) The 6th term is ar 5 ad the 7th term is ar 6 Hece, ar 5 + ar 6 = 1944 Sice r = 3, 43a + 79a = 1944 i.e. 97a = 1944 ad first term, a = = (c) Sum of the 4th to 10th terms iclusive is give by: S 10 3 ( ) ( r ) ( ) ( r ) ( ) ( ) ( ) a r10 1 a r S = = 1 1 (3 1) 3 1 = ( ) ( ) = = = Which term of the series 3, 9, 7,... is ? First term, a = 3, commo ratio, r = 3 ad the th term of a geometric series is: ar 1 Thus, = 33 ( ) from which, 3 1 = = Takig logarithms gives: lg3 1 = lg i.e. ( 1)lg 3 = lg ad 1 = lg lg 3 = 9 from which, = = 10 i.e is the 10th term of the series 3, 9, 7, , Joh Bird

15 3. The value of a lathe origially valued at 3000 depreciates 15% per aum. Calculate its value after four years. The machie is sold whe its value is less tha 550. After how may years is the lathe sold? a = 3000, r = 0.85 ad = 4, hece the value after oe year is: (0.85)(3000) the value after two years is: (0.85) (3000) ad the value after four years is: (0.85) 4 (3000) = Let 550 = (0.85) (3000) from which, = Takig logarithms to base 10 of both sides of the equatio gives: 550 lg lg 0.85 = = lg lg 3000 Hece, umber of years before lathe is sold, = lg 0.85 = = 11 years to the earest whole umber 4. If the populatio of Great Britai is 55 millio ad is decreasig at.4% per aum, what will be the populatio i five years time? GB populatio ow = 55 millio, populatio after oe year = millio Populatio after two years = ( 0.976) 55 millio Hece, populatio after five years = ( 0.976) 5 55 = millio g of a radioactive substace disitegrates at a rate of 3% per aum. How much of the substace is left after 11 years? a = 100, r = 0.97 ad = 11 ad the amout of substace left after 11 years is: (0.97) 11 (100) = g , Joh Bird

16 6. If 50 is ivested at compoud iterest of 6% per aum, determie (a) the value after 15 years, (b) the time, correct to the earest year, it takes to reach 750 (a) First term, a = 50, commo ratio, r = 1.06 Hece, value after 15 years = ar 15 = (50)( 1.06) 15 = (b) Whe 750 is reached, 750 = ar i.e. 750 = 50( 1.06) ad = 1.06 i.e. 3 = 1.06 Takig logarithms gives: lg 3 = lg ( 1.06) = lg1.06 from which, = lg 3 lg1.06 = Hece, it will take 19 years to reach more tha A drillig machie is to have eight speeds ragig from 100 rev/mi to 1000 rev/mi. If the speeds form a geometric progressio, determie their values, each correct to the earest whole umber. First term, a = 100 rev/mi 1000 The 8th term is give by: ar8 1= 1000 from which, r 7 = = 10 ad r = 7 10 = Hece, 1st term is 100 rev/mi d term is ar = (100)(1.3895) = rd term is ar = (100)(1.3895) = th term is ar 3 = (100)(1.3895) 3 = th term is ar 4 = (100)(1.3895) 4 = th term is ar 5 = (100)(1.3895) 5 = th term is ar 6 = (100)(1.3895) 6 = , Joh Bird

17 8th term is ar 7 = (100)(1.3895) 7 = 1000 Hece, correct to the earest whole umbers, the eight speeds are: 100, 139, 193, 68, 373, 518, 70 ad 1000 rev/mi , Joh Bird

18 EXERCISE 83 Page Evaluate: (a) 9 C 6 (b) 3 C 1 (a) 9 C 6 9! 9! = = = = = !(9 6)! 6!3! ( )( ) = 84 (b) 3 3! 3! 3 C1 = = = 1!(3 1)! 1!! 1 ( )( ) = 3. Evaluate: (a) 6 C (b) 8 C 5 (a) 6 C 6! 6! = = = = = 3 5!(6 )!!4! 4 3 ( )( ) = 15 (b) 8 C 5 8! 8! = = = = = 4 7 5!(8 5)! 5!3! ( )( ) = Evaluate: (a) 4 P (b) 7 P 4 (a) 4 4! 4! 4 3 P = = = = 4 3 (4 )!! ( ) = 1 (b) 7 7! 7! P4 = = = = (7 4)! 3! 3 ( ) = Evaluate: (a) 10 P 3 (b) 8 P 5 (a) 10 P3 10! 10! = = = = (10 3)! 7! = 70 (b) 8 8! 8! P5 = = = = = 670 (8 5)! 3! 3 ( ) , Joh Bird