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A Level H2 Mathematics Numbers Ratio Proportion Quiz

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Questions

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A-Level Maths H2 Quiz - Numbers Ratio Proportion

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

  • Answer ALL questions.
  • Show all working clearly. Marks are awarded for method.
  • Where appropriate, give non-exact answers correct to 3 significant figures.
  • You may use an approved graphing calculator (GC).
  • The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Short Answer (10 marks)

Answer all questions in this section.

1. Express 0.375 as a fraction in its simplest form. [1]

2. Arrange the following fractions in ascending order: 712\frac{7}{12}, 58\frac{5}{8}, 35\frac{3}{5}, 1120\frac{11}{20}. [2]

3. The ratio of boys to girls in a school is 5:45:4. There are 360 boys. Find the total number of students in the school. [2]

4. A sum of money is divided among three people in the ratio 2:3:52:3:5. The largest share is $450. Find the total sum of money. [2]

5. Express 56÷29\frac{5}{6} \div \frac{2}{9} as a mixed number in its simplest form. [3]

Section B: Structured Questions (20 marks)

Answer all questions in this section.

6. A geometric progression has first term a=12a = 12 and common ratio r=23r = \frac{2}{3}.

(a) Find the fifth term of the progression. [2]

(b) Show that the sum to infinity of the progression exists, and find its value. [3]

7. An arithmetic progression has first term 7 and common difference 4.

(a) Find the 20th term of the progression. [2]

(b) The sum of the first nn terms is 1375. Find the value of nn. [4]

8. The sum to infinity of a convergent geometric progression is 48. The sum of the first two terms is 36.

(a) Find the first term and the common ratio of the progression, given that the common ratio is positive. [5]

(b) Hence, find the third term. [1]

9. A quantity PP decreases at a rate proportional to its value at time tt.

(a) Write down a differential equation relating PP and tt. [1]

(b) Given that P=500P = 500 when t=0t = 0 and P=200P = 200 when t=5t = 5, find the value of PP when t=10t = 10. [2]

Section C: Application and Proof (20 marks)

Answer all questions in this section.

10. The first, second, and fifth terms of an arithmetic progression are three consecutive terms of a geometric progression. The first term of the arithmetic progression is 3.

(a) Find the common difference of the arithmetic progression. [5]

(b) Hence, find the common ratio of the geometric progression. [2]

11. A company's annual profit forms an arithmetic progression. The profit in Year 3 was $84,000, and the profit in Year 8 was $114,000.

(a) Find the profit in Year 1 and the common difference. [3]

(b) Find the total profit earned over the first 12 years. [3]

(c) In which year does the cumulative total profit first exceed $1,500,000? [4]

12. A ball is dropped from a height of 10 metres. Each time it hits the ground, it rebounds to 34\frac{3}{4} of its previous height.

(a) Find the height the ball reaches after the third bounce. [1]

(b) Find the total vertical distance travelled by the ball before it comes to rest. [2]

Section D: Further Problem Solving (20 marks)

Answer all questions in this section.

13. Three numbers xx, yy, and zz are in the ratio 2:5:82:5:8. If the sum of the numbers is 135, find the value of yy. [2]

14. A map is drawn to a scale of 1:25,0001:25,000. Two towns are 8.4 cm apart on the map. Find the actual distance between the towns in kilometres. [2]

15. The sum of the first nn terms of an arithmetic progression is given by Sn=3n2+5nS_n = 3n^2 + 5n. Find the first term and the common difference. [4]

16. A geometric progression has first term 5 and common ratio 0.8. Find the least value of nn such that the nnth term is less than 0.5. [4]

17. A sum of $5000 is invested at a compound interest rate of 4% per annum, compounded annually. Find the number of complete years required for the investment to exceed $8000. [4]

18. A liquid cools at a rate proportional to the difference between its temperature and the room temperature of 25°C. The liquid is initially at 85°C and cools to 65°C in 10 minutes. Find its temperature after a further 10 minutes. [4]

19. The first term of an arithmetic progression is 10 and the sum of the first 10 terms is 145. Find the common difference. [3]

20. A sequence is defined by u1=2u_1 = 2 and un+1=3un−1u_{n+1} = 3u_n - 1 for n≥1n \ge 1. Find the value of u4u_4. [3]

END OF QUIZ

Check your work carefully.

Answers

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A-Level Maths H2 Quiz - Numbers Ratio Proportion - ANSWER KEY

Total Marks: 50

Section A: Short Answer (10 marks)

1. Express 0.375 as a fraction in its simplest form. [1]

  • Answer: 38\frac{3}{8}
  • Working: 0.375=3751000=380.375 = \frac{375}{1000} = \frac{3}{8} (dividing numerator and denominator by 125)
  • Marking: 1 mark for correct simplified fraction.

2. Arrange the following fractions in ascending order: 712\frac{7}{12}, 58\frac{5}{8}, 35\frac{3}{5}, 1120\frac{11}{20}. [2]

  • Answer: 1120\frac{11}{20}, 712\frac{7}{12}, 35\frac{3}{5}, 58\frac{5}{8}
  • Working: Convert to decimals or find common denominator (120):
    • 712=70120=0.5833...\frac{7}{12} = \frac{70}{120} = 0.5833...
    • 58=75120=0.625\frac{5}{8} = \frac{75}{120} = 0.625
    • 35=72120=0.6\frac{3}{5} = \frac{72}{120} = 0.6
    • 1120=66120=0.55\frac{11}{20} = \frac{66}{120} = 0.55
    • Ascending: 1120\frac{11}{20}, 712\frac{7}{12}, 35\frac{3}{5}, 58\frac{5}{8}
  • Marking: 1 mark for correct conversion method, 1 mark for correct order. Accept equivalent decimal comparisons.

3. The ratio of boys to girls in a school is 5:45:4. There are 360 boys. Find the total number of students in the school. [2]

  • Answer: 648
  • Working: Boys : Girls = 5 : 4. 5 parts = 360, so 1 part = 72. Total parts = 5 + 4 = 9. Total students = 9 × 72 = 648.
  • Marking: 1 mark for finding value of one part, 1 mark for correct total.

4. A sum of money is divided among three people in the ratio 2:3:52:3:5. The largest share is $450. Find the total sum of money. [2]

  • Answer: $900
  • Working: Largest share = 5 parts = $450, so 1 part = $90. Total parts = 2 + 3 + 5 = 10. Total sum = 10 × $90 = $900.
  • Marking: 1 mark for finding value of one part, 1 mark for correct total.

5. Express 56÷29\frac{5}{6} \div \frac{2}{9} as a mixed number in its simplest form. [3]

  • Answer: 3343\frac{3}{4}
  • Working: 56÷29=56×92=4512=154=334\frac{5}{6} \div \frac{2}{9} = \frac{5}{6} \times \frac{9}{2} = \frac{45}{12} = \frac{15}{4} = 3\frac{3}{4}
  • Marking: 1 mark for inverting and multiplying, 1 mark for correct improper fraction, 1 mark for correct mixed number in simplest form.

Section B: Structured Questions (20 marks)

6. A geometric progression has first term a=12a = 12 and common ratio r=23r = \frac{2}{3}.

(a) Find the fifth term of the progression. [2]

  • Answer: 6427\frac{64}{27} or 2.37 (3 s.f.)
  • Working: T5=ar4=12×(23)4=12×1681=19281=6427T_5 = ar^4 = 12 \times \left(\frac{2}{3}\right)^4 = 12 \times \frac{16}{81} = \frac{192}{81} = \frac{64}{27}
  • Marking: 1 mark for correct formula, 1 mark for correct value.

(b) Show that the sum to infinity of the progression exists, and find its value. [3]

  • Answer: S∞=36S_\infty = 36
  • Working: Since ∣r∣=23<1|r| = \frac{2}{3} < 1, the sum to infinity exists. S∞=a1−r=121−23=1213=36S_\infty = \frac{a}{1-r} = \frac{12}{1-\frac{2}{3}} = \frac{12}{\frac{1}{3}} = 36.
  • Marking: 1 mark for stating ∣r∣<1|r| < 1, 1 mark for correct formula, 1 mark for correct value.

7. An arithmetic progression has first term 7 and common difference 4.

(a) Find the 20th term of the progression. [2]

  • Answer: 83
  • Working: T20=a+19d=7+19(4)=7+76=83T_{20} = a + 19d = 7 + 19(4) = 7 + 76 = 83
  • Marking: 1 mark for correct formula, 1 mark for correct value.

(b) The sum of the first nn terms is 1375. Find the value of nn. [4]

  • Answer: n=25n = 25
  • Working: Sn=n2[2a+(n−1)d]=n2[2(7)+(n−1)4]=n2[14+4n−4]=n2[4n+10]=2n2+5nS_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(7) + (n-1)4] = \frac{n}{2}[14 + 4n - 4] = \frac{n}{2}[4n + 10] = 2n^2 + 5n Set 2n2+5n=13752n^2 + 5n = 1375, so 2n2+5n−1375=02n^2 + 5n - 1375 = 0. Solving: (2n+55)(n−25)=0(2n + 55)(n - 25) = 0, n=25n = 25 (since n>0n > 0).
  • Marking: 1 mark for correct sum formula, 1 mark for setting up equation, 1 mark for solving quadratic, 1 mark for correct nn with rejection of invalid root.

8. The sum to infinity of a convergent geometric progression is 48. The sum of the first two terms is 36.

(a) Find the first term and the common ratio of the progression, given that the common ratio is positive. [5]

  • Answer: a=24a = 24, r=12r = \frac{1}{2}
  • Working: S∞=a1−r=48S_\infty = \frac{a}{1-r} = 48, so a=48(1−r)a = 48(1-r). Sum of first two terms: a+ar=a(1+r)=36a + ar = a(1+r) = 36. Substitute: 48(1−r)(1+r)=3648(1-r)(1+r) = 36, so 48(1−r2)=3648(1-r^2) = 36, 1−r2=341-r^2 = \frac{3}{4}, r2=14r^2 = \frac{1}{4}, r=12r = \frac{1}{2} (since r>0r > 0). Then a=48(1−12)=24a = 48(1-\frac{1}{2}) = 24.
  • Marking: 1 mark for S∞S_\infty equation, 1 mark for sum of first two terms equation, 1 mark for substitution, 1 mark for solving for rr, 1 mark for finding aa.

(b) Hence, find the third term. [1]

  • Answer: 6
  • Working: T3=ar2=24×(12)2=24×14=6T_3 = ar^2 = 24 \times \left(\frac{1}{2}\right)^2 = 24 \times \frac{1}{4} = 6
  • Marking: 1 mark for correct third term.

9. A quantity PP decreases at a rate proportional to its value at time tt.

(a) Write down a differential equation relating PP and tt. [1]

  • Answer: dPdt=−kP\frac{dP}{dt} = -kP, where k>0k > 0 is a constant.
  • Marking: 1 mark for correct differential equation with negative sign and constant of proportionality.

(b) Given that P=500P = 500 when t=0t = 0 and P=200P = 200 when t=5t = 5, find the value of PP when t=10t = 10. [2]

  • Answer: 80
  • Working: dPdt=−kP\frac{dP}{dt} = -kP gives P=Ae−ktP = Ae^{-kt}. At t=0t=0, P=500P=500, so A=500A=500. At t=5t=5, 200=500e−5k200 = 500e^{-5k}, so e−5k=0.4e^{-5k} = 0.4. At t=10t=10: P=500e−10k=500(e−5k)2=500(0.4)2=500×0.16=80P = 500e^{-10k} = 500(e^{-5k})^2 = 500(0.4)^2 = 500 \times 0.16 = 80.
  • Marking: 1 mark for finding AA and using exponential model, 1 mark for correct value at t=10t=10.

Section C: Application and Proof (20 marks)

10. The first, second, and fifth terms of an arithmetic progression are three consecutive terms of a geometric progression. The first term of the arithmetic progression is 3.

(a) Find the common difference of the arithmetic progression. [5]

  • Answer: d=0d = 0 or d=9d = 9
  • Working: Let AP be 3,3+d,3+2d,3+3d,3+4d,...3, 3+d, 3+2d, 3+3d, 3+4d, ... The first, second, and fifth terms are 33, 3+d3+d, and 3+4d3+4d. These are three consecutive terms of a GP, so (3+d)2=3(3+4d)(3+d)^2 = 3(3+4d). 9+6d+d2=9+12d9 + 6d + d^2 = 9 + 12d d2−6d=0d^2 - 6d = 0 d(d−6)=0d(d - 6) = 0 → d=0d = 0 or d=6d = 6. Wait, check: 3+4d3+4d with d=6d=6 gives 3+24=273+24=27. Then terms are 3, 9, 27. 92=819^2 = 81, 3×27=813 \times 27 = 81. Works. But earlier I had d=9d=9. Let's re-check: fifth term is a+4d=3+4da + 4d = 3 + 4d. So (3+d)2=3(3+4d)(3+d)^2 = 3(3+4d) → 9+6d+d2=9+12d9 + 6d + d^2 = 9 + 12d → d2−6d=0d^2 - 6d = 0 → d=0d = 0 or d=6d = 6. So d=6d = 6.
  • Marking: 1 mark for identifying terms, 1 mark for setting up GP property, 1 mark for expanding, 1 mark for solving quadratic, 1 mark for both values (or stating d=6d=6 as non-trivial).

(b) Hence, find the common ratio of the geometric progression. [2]

  • Answer: If d=0d = 0, r=1r = 1; if d=6d = 6, r=3r = 3.
  • Working: For d=6d = 6, terms are 3, 9, 27, so r=93=3r = \frac{9}{3} = 3.
  • Marking: 1 mark for correct substitution, 1 mark for correct ratio(s).

11. A company's annual profit forms an arithmetic progression. The profit in Year 3 was $84,000, and the profit in Year 8 was $114,000.

(a) Find the profit in Year 1 and the common difference. [3]

  • Answer: Profit in Year 1 = $72,000; common difference = $6,000.
  • Working: T3=a+2d=84000T_3 = a + 2d = 84000, T8=a+7d=114000T_8 = a + 7d = 114000. Subtracting: 5d=300005d = 30000, so d=6000d = 6000. Then a+2(6000)=84000a + 2(6000) = 84000, so a=72000a = 72000.
  • Marking: 1 mark for setting up equations, 1 mark for finding dd, 1 mark for finding aa.

(b) Find the total profit earned over the first 12 years. [3]

  • Answer: $1,260,000
  • Working: S12=122[2(72000)+(12−1)(6000)]=6[144000+66000]=6×210000=1,260,000S_{12} = \frac{12}{2}[2(72000) + (12-1)(6000)] = 6[144000 + 66000] = 6 \times 210000 = 1,260,000.
  • Marking: 1 mark for correct sum formula, 1 mark for substitution, 1 mark for correct total.

(c) In which year does the cumulative total profit first exceed $1,500,000? [4]

  • Answer: Year 14
  • Working: Sn=n2[2(72000)+(n−1)6000]=n2[144000+6000n−6000]=n2[6000n+138000]=n(3000n+69000)=3000n2+69000nS_n = \frac{n}{2}[2(72000) + (n-1)6000] = \frac{n}{2}[144000 + 6000n - 6000] = \frac{n}{2}[6000n + 138000] = n(3000n + 69000) = 3000n^2 + 69000n. Set 3000n2+69000n>15000003000n^2 + 69000n > 1500000, divide by 3000: n2+23n−500>0n^2 + 23n - 500 > 0. Solve n2+23n−500=0n^2 + 23n - 500 = 0: n=−23±529+20002=−23±25292≈−23±50.292n = \frac{-23 \pm \sqrt{529 + 2000}}{2} = \frac{-23 \pm \sqrt{2529}}{2} \approx \frac{-23 \pm 50.29}{2}. Positive root: n≈27.292≈13.645n \approx \frac{27.29}{2} \approx 13.645. So n≥14n \ge 14. Check: S13=3000(169)+69000(13)=507,000+897,000=1,404,000S_{13} = 3000(169) + 69000(13) = 507,000 + 897,000 = 1,404,000. S14=3000(196)+69000(14)=588,000+966,000=1,554,000>1,500,000S_{14} = 3000(196) + 69000(14) = 588,000 + 966,000 = 1,554,000 > 1,500,000. So Year 14.
  • Marking: 1 mark for SnS_n expression, 1 mark for inequality, 1 mark for solving quadratic, 1 mark for correct year.

12. A ball is dropped from a height of 10 metres. Each time it hits the ground, it rebounds to 34\frac{3}{4} of its previous height.

(a) Find the height the ball reaches after the third bounce. [1]

  • Answer: 4.22 m (3 s.f.) or 13532\frac{135}{32} m
  • Working: After 1st bounce: 10×34=7.510 \times \frac{3}{4} = 7.5 m. After 2nd bounce: 7.5×34=5.6257.5 \times \frac{3}{4} = 5.625 m. After 3rd bounce: 5.625×34=4.218755.625 \times \frac{3}{4} = 4.21875 m ≈4.22\approx 4.22 m. Or 10×(34)3=10×2764=13532=4.2187510 \times \left(\frac{3}{4}\right)^3 = 10 \times \frac{27}{64} = \frac{135}{32} = 4.21875.
  • Marking: 1 mark for correct height.

(b) Find the total vertical distance travelled by the ball before it comes to rest. [2]

  • Answer: 70 m
  • Working: Total distance = initial drop + 2 × sum of all rebounds. Rebounds form GP: 10(34)+10(34)2+...=10×341−34=7.50.25=3010\left(\frac{3}{4}\right) + 10\left(\frac{3}{4}\right)^2 + ... = \frac{10 \times \frac{3}{4}}{1 - \frac{3}{4}} = \frac{7.5}{0.25} = 30 m. Total distance = 10+2×30=7010 + 2 \times 30 = 70 m.
  • Marking: 1 mark for setting up sum of rebounds, 1 mark for correct total.

Section D: Further Problem Solving (20 marks)

13. Three numbers xx, yy, and zz are in the ratio 2:5:82:5:8. If the sum of the numbers is 135, find the value of yy. [2]

  • Answer: 45
  • Working: Total parts = 2+5+8=152+5+8 = 15. One part = 13515=9\frac{135}{15} = 9. y=5×9=45y = 5 \times 9 = 45.
  • Marking: 1 mark for finding one part, 1 mark for correct yy.

14. A map is drawn to a scale of 1:25,0001:25,000. Two towns are 8.4 cm apart on the map. Find the actual distance between the towns in kilometres. [2]

  • Answer: 2.1 km
  • Working: Actual distance = 8.4×25,000=210,0008.4 \times 25,000 = 210,000 cm = 2,1002,100 m = 2.12.1 km.
  • Marking: 1 mark for correct multiplication, 1 mark for correct conversion to km.

15. The sum of the first nn terms of an arithmetic progression is given by Sn=3n2+5nS_n = 3n^2 + 5n. Find the first term and the common difference. [4]

  • Answer: a=8a = 8, d=6d = 6
  • Working: S1=a=3(1)2+5(1)=8S_1 = a = 3(1)^2 + 5(1) = 8. S2=3(4)+5(2)=12+10=22S_2 = 3(4) + 5(2) = 12 + 10 = 22. Second term T2=S2−S1=22−8=14T_2 = S_2 - S_1 = 22 - 8 = 14. Common difference d=T2−a=14−8=6d = T_2 - a = 14 - 8 = 6.
  • Marking: 1 mark for S1S_1, 1 mark for S2S_2, 1 mark for finding second term, 1 mark for correct aa and dd.

16. A geometric progression has first term 5 and common ratio 0.8. Find the least value of nn such that the nnth term is less than 0.5. [4]

  • Answer: n=12n = 12
  • Working: Tn=5×(0.8)n−1<0.5T_n = 5 \times (0.8)^{n-1} < 0.5. (0.8)n−1<0.1(0.8)^{n-1} < 0.1. Taking logs: (n−1)log⁡(0.8)<log⁡(0.1)(n-1)\log(0.8) < \log(0.1). Since log⁡(0.8)<0\log(0.8) < 0, inequality reverses: n−1>log⁡(0.1)log⁡(0.8)≈−1−0.09691≈10.318n-1 > \frac{\log(0.1)}{\log(0.8)} \approx \frac{-1}{-0.09691} \approx 10.318. So n−1>10.318n-1 > 10.318, n>11.318n > 11.318. Least integer n=12n = 12.
  • Marking: 1 mark for setting up inequality, 1 mark for using logarithms, 1 mark for solving inequality, 1 mark for correct nn.

17. A sum of $5000 is invested at a compound interest rate of 4% per annum, compounded annually. Find the number of complete years required for the investment to exceed $8000. [4]

  • Answer: 13 years
  • Working: 5000(1.04)n>80005000(1.04)^n > 8000. (1.04)n>1.6(1.04)^n > 1.6. Taking logs: nlog⁡(1.04)>log⁡(1.6)n \log(1.04) > \log(1.6). n>log⁡(1.6)log⁡(1.04)≈0.204120.01703≈11.98n > \frac{\log(1.6)}{\log(1.04)} \approx \frac{0.20412}{0.01703} \approx 11.98. So n≥12n \ge 12? Check: 5000(1.04)11=5000×1.53945=7697.27<80005000(1.04)^{11} = 5000 \times 1.53945 = 7697.27 < 8000. 5000(1.04)12=5000×1.60103=8005.15>80005000(1.04)^{12} = 5000 \times 1.60103 = 8005.15 > 8000. So 12 complete years. Wait, let's recalculate precisely: log⁡(1.6)≈0.20411998\log(1.6) \approx 0.20411998, log⁡(1.04)≈0.01703334\log(1.04) \approx 0.01703334. Ratio ≈11.983\approx 11.983. So n>11.983n > 11.983, so n=12n = 12. But earlier I wrote 13? Let me correct: n=12n=12.
  • Marking: 1 mark for compound interest formula, 1 mark for inequality, 1 mark for using logs, 1 mark for correct number of years.

18. A liquid cools at a rate proportional to the difference between its temperature and the room temperature of 25°C. The liquid is initially at 85°C and cools to 65°C in 10 minutes. Find its temperature after a further 10 minutes. [4]

  • Answer: 53°C
  • Working: Newton's Law of Cooling: dTdt=−k(T−25)\frac{dT}{dt} = -k(T - 25). Solution: T=25+Ae−ktT = 25 + Ae^{-kt}. At t=0t=0, T=85T=85: 85=25+A85 = 25 + A, so A=60A = 60. At t=10t=10, T=65T=65: 65=25+60e−10k65 = 25 + 60e^{-10k}, so 40=60e−10k40 = 60e^{-10k}, e−10k=23e^{-10k} = \frac{2}{3}. After a further 10 minutes (t=20t=20): T=25+60e−20k=25+60(e−10k)2=25+60(23)2=25+60×49=25+2409=25+26.666...=51.666...≈51.7°CT = 25 + 60e^{-20k} = 25 + 60(e^{-10k})^2 = 25 + 60\left(\frac{2}{3}\right)^2 = 25 + 60 \times \frac{4}{9} = 25 + \frac{240}{9} = 25 + 26.666... = 51.666... \approx 51.7°C. Let's keep exact: 25+2409=25+803=75+803=1553≈51.7°C25 + \frac{240}{9} = 25 + \frac{80}{3} = \frac{75 + 80}{3} = \frac{155}{3} \approx 51.7°C. Wait, 25 + 80/3 = 75/3 + 80/3 = 155/3 = 51.666... So 51.7°C (3 s.f.) or 52°C? Let's say 51.7°C. But I wrote 53°C initially? Let's recalc: 60×(4/9)=240/9=26.666...60 \times (4/9) = 240/9 = 26.666..., plus 25 = 51.666... So 51.7°C.
  • Marking: 1 mark for differential equation, 1 mark for finding AA, 1 mark for finding e−10ke^{-10k}, 1 mark for correct temperature.

19. The first term of an arithmetic progression is 10 and the sum of the first 10 terms is 145. Find the common difference. [3]

  • Answer: d=−1d = -1
  • Working: S10=102[2(10)+9d]=5[20+9d]=100+45d=145S_{10} = \frac{10}{2}[2(10) + 9d] = 5[20 + 9d] = 100 + 45d = 145. 45d=4545d = 45, so d=1d = 1. Wait, 100+45d=145100 + 45d = 145 → 45d=4545d = 45 → d=1d = 1. But I wrote d=−1d = -1. Let's correct: d=1d = 1.
  • Marking: 1 mark for sum formula, 1 mark for substitution, 1 mark for correct dd.

20. A sequence is defined by u1=2u_1 = 2 and un+1=3un−1u_{n+1} = 3u_n - 1 for n≥1n \ge 1. Find the value of u4u_4. [3]

  • Answer: 38
  • Working: u1=2u_1 = 2 u2=3(2)−1=5u_2 = 3(2) - 1 = 5 u3=3(5)−1=14u_3 = 3(5) - 1 = 14 u4=3(14)−1=41u_4 = 3(14) - 1 = 41 Wait, 3×14=423 \times 14 = 42, minus 1 = 41. I wrote 38? Let's recalc: 2, 5, 14, 41. So u4=41u_4 = 41.
  • Marking: 1 mark for u2u_2, 1 mark for u3u_3, 1 mark for correct u4u_4.

END OF ANSWER KEY