IB Math SL Arithmetic Geometric Sequences Series Review

IB Math SL 1. 2. 3. 4. 5. 6.

Arithmetic & Geometric Sequences/Series Review

Find the sum of the arithmetic series 17 + 27 + 37 +...+ 417. An arithmetic series has five terms. The first term is 2 and the last term is 32. Find the sum of the series. 8  16  ... Find the sum of the infinite geometric series 2  4  3 9 27 81 Gwendolyn added the multiples of 3, from 3 to 3750 and found that 3 + 6 + 9 + … + 3750 = s. Calculate s. In an arithmetic sequence, the first term is 5 and the fourth term is 40. Find the second term. The Acme insurance company sells two savings plans, Plan A and Plan B. For Plan A, an investor starts with an initial deposit of $1000 and increases this by $80 each month, so that in the second month, the deposit is $1080, the next month it is $1160 and so on. For Plan B, the investor again starts with $1000 and each month deposits 6% more than the previous month. (a) Write down the amount of money invested under Plan B in the second and third months. (2)

Give your answers to parts (b) and (c) correct to the nearest dollar. (b) Find the amount of the 12th deposit for each Plan. (4)

(c)

Find the total amount of money invested during the first 12 months (i) under Plan A; (2)

(ii)

under Plan B. (2) (Total 10 marks)

$1000 is invested at the beginning of each year for 10 years. The rate of interest is fixed at 7.5% per annum. Interest is compounded annually. Calculate, giving your answers to the nearest dollar (a) how much the first $1000 is worth at the end of the ten years; (b) the total value of the investments at the end of the ten years. 8. Each day a runner trains for a 10 km race. On the first day she runs 1000 m, and then increases the distance by 250 m on each subsequent day. (a) On which day does she run a distance of 10 km in training? (b) What is the total distance she will have run in training by the end of that day? Give your answer exactly. 9. The first three terms of an arithmetic sequence are 7, 9.5, 12. st (a) What is the 41 term of the sequence? (b) What is the sum of the first 101 terms of the sequence? 10. Portable telephones are first sold in the country Cellmania in 1990. During 1990, the number of units sold is 160. In 1991, the number of units sold is 240 and in 1992, the number of units sold is 360. In 1993 it was noticed that the annual sales formed a geometric sequence with first term 160, the 2nd and 3rd terms being 240 and 360 respectively. (a) What is the common ratio of this sequence?

7.

(1)

Assume that this trend in sales continues. (b) How many units will be sold during 2002? (3)

(c)

In what year does the number of units sold first exceed 5000? (4)

Between 1990 and 1992, the total number of units sold is 760. (d) What is the total number of units sold between 1990 and 2002? (2)

During this period, the total population of Cellmania remains approximately 80 000. (e) Use this information to suggest a reason why the geometric growth in sales would not continue. (1) (Total 11 marks)

11. In an arithmetic sequence, the first term is –2, the fourth term is 16, and the nth term is 11 998. (a) (b)

Find the common difference d. Find the value of n.

12. Ashley and Billie are swimmers training for a competition. (a)

Ashley trains for 12 hours in the first week. She decides to increase the amount of time she spends training by 2 hours each week. Find the total number of hours she spends training during the first 15 weeks.

(b)

Billie also trains for 12 hours in the first week. She decides to train for 10% longer each week than the previous week. (i) Show that in the third week she trains for 14.52 hours. (ii) Find the total number of hours she spends training during the first 15 weeks.

(c)

In which week will the time Billie spends training first exceed 50 hours?

(3)

(4) (4) (Total 11 marks)

13. Arturo goes swimming every week. He swims 200 metres in the first week. Each week he swims 30

metres more than the previous week. He continues for one year (52 weeks). (a) How far does Arturo swim in the final week? (b) How far does he swim altogether? 14. A company offers its employees a choice of two salary schemes A and B over a period of 10 years. Scheme A offers a starting salary of $11 000 in the first year and then an annual increase of $400 per year. (a) (i) Write down the salary paid in the second year and in the third year. (ii) Calculate the total (amount of) salary paid over ten years.

(3)

Scheme B offers a starting salary of $10 000 dollars in the first year and then an annual increase of 7 % of the previous year’s salary. (b) (i) Write down the salary paid in the second year and in the third year. (ii) Calculate the salary paid in the tenth year. (4)

(c)

Arturo works for n complete years under scheme A. Bill works for n complete years under scheme B. Find the minimum number of years so that the total earned by Bill exceeds the total earned by Arturo. (4) (Total 11 marks)

15. A theatre has 20 rows of seats. There are 15 seats in the first row, 17 seats in the second row, and each

successive row of seats has two more seats in it than the previous row. th (a) Calculate the number of seats in the 20 row. (b) Calculate the total number of seats. 16. A sum of $5 000 is invested at a compound interest rate of 6.3% per annum. (a) Write down an expression for the value of the investment after n full years. (b) What will be the value of the investment at the end of five years? (c) The value of the investment will exceed $10 000 after n full years, (i) Write down an inequality to represent this information. (ii) Calculate the minimum value of n. 17. Let Sn be the sum of the first n terms of an arithmetic sequence, whose first three terms are u1, u2 and u3. It is known that S1 = 7, and S2 = 18. (a) Write down u1. (b) Calculate the common difference of the sequence. (c) Calculate u4.

IB Math SL 1.

Arithmetic & Geometric Sequences/Series Review Solutions

17 + 27 + 37 + ... + 417 17 + (n – 1)10 = 417 10(n – 1) = 400 n = 41 41 S41 = (2(17) + 40(10)) 2 = 41(17 + 200) = 8897 OR 41 S41 = (17 + 417) 2 41 = (434) 2 = 8897

(M1) (A1) (M1) (A1) (M1)

(A1) [4]

2.

5 S5 = {2 + 32} 2 S5 = 85 OR a = 2, a + 4d = 32  4d = 30 d = 7.5 5 S5 = (4 + 4(7.5)) 2 5 = (4 + 30) 2 S5 = 85

(M1)(A1)(A1) (A1) (M1) (A1) (M1)

(A1) [4]

3. S=

u1  1 r

2 3

 2 1     3 2 3 =  3 5 2 = 5

(M1)(A1)

(A1) (A1) (C4) [4]

4.

Arithmetic sequence d = 3 (may be implied) n = 1250 1250 1250   (6  1249  3)  S = (3 + 3750)  or S  2 2   = 2 345 625

(M1)(A1) (A2) (M1) (A1) (C6) [6]

5.

a=5 a + 3d = 40 (may be implied) 35 d= 3 35 T2 = 5 + 3 2 50 = 16 or or 16.7 (3 s.f.) 3 3

(M1) (A1) (A1) (A1) (C4) [4]

6.

(a)

2

Plan A: 1000, 1080, 1160... Plan B: 1000, 1000(1.06), 1000(1.06) … 2nd month: $ 1060, 3rd month: $ 1123.60 (A1)(A1) (b) For Plan A, T12 = a + 11d = 1000 + 11(80) (M1) = $ 1880 (A1) 11 For Plan B, T12 = 1000(1.06) (M1) = $ 1898 (to the nearest dollar) (A1) 12 (c) (i) For Plan A, S12 = [2000 + 11(80)] (M1) 2 = 6(2880) = $17280 (to the nearest dollar) (A1) 12 1000(1.06  1) (ii) For Plan B, S12 = (M1) 1.06  1 = $ 16870 (to the nearest dollar) (A1)

2

4

4 [10]

7.

(a)

10

$ 1000 × 1.075 = $ 2061 (nearest dollar) 10 9 (b) 1000(1.075 + 1.075 + ... + 1.075) 1000(1.075)(1.07510  1) = 1.075  1 = $ 15 208 (nearest dollar)

(A1) (C1) (M1) (M1) (A1) (C3) [4]

8.

(a)

a1 = 1000, an = 1000 + (n – 1)250 = 10 000 10 000  1000 n= + 1 = 37. 250 She runs 10 km on the 37th day. 37 (b) S37 = (1000 + 10 000) 2 She has run a total of 203.5 km

(M1)

u1 = 7, d = 2.5 u41 = u1 + (n – 1)d = 7 + (41 – 1)2.5 = 107 n (b) S101 = [2u1 + (n – 1)d] 2 101 = [2(7) + (101 – 1)2.52] 2

(M1)

(A1) (M1) (A1) [4]

9.

(a)

(A1) (C2)

(M1) 101(264) = 2 = 13332 (A1) (C2) [4]

10.

(a)

360 240 3   = 1.5 240 160 2 th (b) 2002 is the 13 year.(M1) 13–1 u13 = 160(1.5) (M1) = 20759 (Accept 20760 or 20800.) (A1) r=

(c)

3 n–1 5000 = 160(1.5)

(A1)

1

5000 n–1 = (1.5) (M1) 160  5000   = (n – 1)log1.5 (M1) log   160   5000  log  n–1=  160  = 8.49 (A1) log 1.5 th

 n = 9.49  10 year  1999 (A1) OR 3 Using a gdc with u1 = 160, uk+1 = uk, u9 = 4100, u10 = 6150 (M2) 2



1999 (G2)

4 (d)

11. (a)

 1.513  1    1.5  1 

S13 = 160 

(M1)

= 61958 (Accept 61960 or 62 000.) (A1)2 (e) Nearly everyone would have bought a portable telephone so there would be fewer people left wanting to buy one. (R1) u4 = ul + 3d or 16 = –2 +3d (M1) d= (b)

16 –  – 2  3

(M1)

=6 un = ul + (n – 1) 6 or 11998 = –2 + (n – l)6

(A1) (C3) (M1)

n=

(A1)

11998  2 1 6

= 2001

(A1) (C3) [6]

12. (a)

Ashley AP

(b)

12 + 14 + 16 + ... to 15 terms

(M1)

15 S15 = [2(12) + 14(2)] 2

(M1)

= 15 × 26 = 390 hours

(A1)

3

Billie 2

(i)

GP 12, 12(1.1), 12(1.1) … 2 In week 3, 12(1.1) = 14.52 hours

(ii)

S15 =

12[1.1 – 1] 1.1 – 1

(M1) (A1) (AG)

15

(c)

= 381 hours (3 s.f.) 12 (1.1) > 50 n–1

(1.1)

n–1

>

50 12

(n – 1) ln 1.1 > ln

(M1) (A1) (M1) (A1)

50 12

50 n–1> 12 ln 1.1 ln

n – 1 > 14.97 n > 15.97  Week 16 n–1 OR 12(1.1) > 50 By trial and error

(A1)

(A1) (M1)

4

14

15

12(1.1) = 45.6, 12(1.1) =50.1  n – l = 15  n = 16 (Week 16)

(A1) (A1) (A1)

4 [11]

13. Arithmetic sequence

(M1) (A1) (M1) (A1) (C3)

a = 200 d = 30 (a) Distance in final week = 200 + 51 × 30 = 1730 m (b)

Total distance =

52 [2.200 + 51.30] 2

(M1)

= 50180m (A1) (C3) Note: Penalize once for absence of units i.e. award A0 the first time units are omitted, A1 the next time. [6]

14. (a)

(i)

$ 11 400, $ 11800

(ii)

(A1)

Total salary  (A1)

1

10 (2 11000  9  400) 2

 $ 128 000

(A1) (b)

(i)

2 $ 10 700, $ 11 449

(ii)

10 year salary  10 000(1.07)

(A1)(A1)

th

(A1)

9

 $18 384.59 or $ 18 400 or $ 18 385

(A1) 4 (c)

EITHER Scheme A

SA 

n  2  11000  (n  1) 400  2

(A1) Scheme B

SB 

10 000(1.07 n  1) 1.07  1

(A1) Solving SB  SA (accept S B  SA , giving n  6.33 ) (may be implied) (M1) Minimum value of n is 7 years. (A1) OR Using trial and error (M1) 6 years 7 years

Arturo $ 72 000 $ 85 400

Bill $ 71532.91 $ 86 540.21

(A1)(A1) Note: Award (A1) for both values for 6 years, and (A1) for both values for 7 years. Therefore, minimum number of years is 7. (A1)

(N2) 4 [11]

15. (a)

Recognizing an AP u1 =15 d = 2 n = 20 substituting into u20 = 15 + (20 –1) × 2 = 53 (that is, 53 seats in the 20th row) (b) Substituting into S20= 20 (2(15) + (20–1)2) (or into 20 (15 + 53)) 2 2

(M1) (A1) M1 A1

4

M1 = 680 (that is, 680 seats in total) A1 2 [6]

16. (a)

n

5 000(1.063) 5 (b) Value = $5 000(1.063) (= $6 786.3511...) = $ 6 790 to 3 s.f. (Accept $ 6 786, or $ 6 786.35)

A1

1

A1

1

A1

(c)

1 (i) (ii)

n

n

5000(1.063) > 10000 or (1.063) > 2 Attempting to solve the inequality «log (1.063) > log 2 (M1) n > 11.345... (A1) 12 years A1 3

Note: Candidates are likely to use TABLE or LIST on a GDC to find n. A good way of communicating this is suggested below. x Let y = 1.063 (M1) When x = 11, y = 1.9582, when x = 12, y = 2.0816 (A1) x = 12 i.e. 12 years A1 3 [6]

17. (a)

u1  S1  7 u2  S2  u1  18  7 (b)  11 d  11  7

(c)

(M1) 4 u4  u1  (n  1) d  7  3(4) u4  19 (A1)

(A1) (C1) (A1)

(A1) (C3) (M1)

(C2) [6]