Solutions — Full Answer Key

Mathematics

Year 8 · 8.6 Patterns

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze

1.17, 20; rule: add 3 each time

2.41

3.(a) 13 (b) Add 3 each time

4.16

5.5, 7

6.7, 9, 11

7.Add 3 each pattern. (

T_n = 3n + 2

-4

9.Yes —

n = 15

10.No —

n = 12.67

(not a whole number).

Silver

11.

T_n = 3n + 1

12.98

13.

T_n = 2n + 1

14.

T_n = -3n + 13

(or

13 - 3n

)

15.

n = 19

16.(a)

d = 4

(b)

T_1 = 3

(c)

T_n = 4n - 1

17.

T_n = n^2

18.21 and 28

19.

T_n = n(n+1)/2

20.

T_n = T_{n-1} + T_{n-2}

;

T_8 = 21

Gold

21.(a) 50 (b) 17 tables

22.

P = 46

23.(a)

T_n = 2n + 1

(b) pattern 25

24.

T_n = 4n + 3

;

T_{50} = 203

25.(a)

T_n = 3n + 1

(b) 33 squares; 0 left

26.

T_n = 3n - 4

27.

T_n = 4n + 1

28.

T_n = n^2 + 4

29.120

30.0, 2, 20, 90

Platinum

31.(a)

T_n = n^2

. (b) Odd numbers 3, 5, 7, … (c)

T_{20} = 400

32.

T_n = 5n - 7

;

n = 22

33.

P_1 = 1, P_2 = 5, P_3 = 12

;

P_{10} = 145

34.1683

35.(a)

T_5 = a + 4d

(b)

a = 1, d = 6

. Wait recompute →

a = 23 - 4 \times 6 = -1

. So

a = -1

d = 6

36.(a)

A_n = 3n + 2

B_n = 5n - 3

(b)

n = 5/2

— never (no whole-number solution)

37.

H_6 = 66, H_{10} = 190

38.(a) 3, 6, 12, 24, 48 (b)

T_n = 3 \times 2^{n-1}

39.(a)

n^2

(b)

1 + 4 + 9 + 16 + 25 = 55

(c)

S_n = \tfrac{n(n+1)(2n+1)}{6}

40.(a) 1, 6, 15 (b) Second differences = 4 (c) No (

n \approx 8.1

)

Pack B — Answers

Bronze

1.27, 33; rule: add 6 each time

2.38

3.(a) 17 (b) Add 4 each time

4.16

5.5, 8

6.2, 5, 8

7.Add 4. (

T_n = 4n + 3

-7

9.Yes —

n = 11

10.Yes —

n = 7

Silver

11.

T_n = 4n + 1

12.5

13.

T_n = 3n + 1

14.

T_n = 5n + 3

15.

n = 32

16.(a)

d = 4

(b)

T_1 = 3

(c)

T_n = 4n - 1

17.Same.

18.36 and 45

19.Same.

20.

T_{10} = 55

Gold

21.(a) 38 (b) 13 tables

22.

P = 52

23.(a)

T_n = 2n + 1

(b) pattern 49

24.

T_n = 4n + 2

;

T_{50} = 202

25.(a) Same (b) 26 squares, 1 stick left

26.

T_n = 3n - 8

27.

T_n = 5n - 11

28.

T_n = n^2 + 2

29.168

30.Same.

Platinum

31.(a)

T_n = n^3

. (b) 7, 19, 37, … (c)

T_{10} = 1000

32.

T_n = 5n - 8

;

n = 22

33.Same.

34.1050

35.(a)

T_4 = a + 3d

(b)

a = 2, d = 3

36.(a)

A_n = 4n - 1

B_n = 5n - 4

(b)

n = 3

—

A_3 = B_3 = 11

37.Same.

38.(a) 2, 6, 18, 54, 162 (b)

T_n = 2 \times 3^{n-1}

39.Same.

40.(a) 4, 14, 30 (b) Second diff = 6 (c) Test directly

Problem-solving — Worked Solutions

1Problem 1

Answer

(a) 1→8, 2→14, 3→20, 4→26 (b)

M = 6n + 2

Full working

(a) Each new table loses one end seat (becomes inner join) but gains 6 side seats — net +6. Values: 1 → 8, 2 → 14, 3 → 20, 4 → 26.

(b) Common difference 6, first term 8:

M = 6n + 2

. Check

n=1

6+2 = 8

✓.

(c)

M = 6(8) + 2 = 50

.

(d) Adding 6 each time: 8, 14, 20, 26, 32, 38, 44, 50 ✓.

Justification: 2 end seats + 6n side seats =

2 + 6n

2Problem 2

Answer

(a)

P = 2an + 2b

(b) Part One:

P = 6n + 2

✓; Part Two:

P = 8n + 4

(c)

P = 46

, not 40

Full working

(a) Each table has

a

side seats on each of 2 sides →

2an

total side seats. Only first & last tables have end seats →

2b

. So

P = 2an + 2b

.

(b) Part One

(a=3, b=1)

P = 6n + 2

✓. Part Two

(a=4, b=2)

P = 8n + 4

.

(c)

P = 2 \times 5 \times 4 + 2 \times 3 = 40 + 6 = 46

. Mr. Packer is **wrong** — 46 people, not 40.

3Problem 3

Answer

(a) 4 (b) 5 (c)

T_n = 4n + 1

(d)

n = 25

Full working

(a) From term 4 to term 10 is 6 steps with a change of 24. Common difference =

24/6 = 4

.

(b)

T_1 + 3d = 17 \Rightarrow T_1 = 5

.

(c)

T_n = 5 + 4(n-1) = 4n + 1

.

(d)

4n + 1 = 101 \Rightarrow n = 25

4Problem 4

Answer

(a)

T_n = n^2

(b) Differences 3, 5, 7 — odd numbers (c)

(n+1)^2 - n^2 = 2n+1

(d)

T_{50} = 2401 + 99 = 2500

Full working

(a)

T_n = n^2

.

(b) Differences 3, 5, 7 — consecutive odd numbers.

(c)

(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1

.

(d)

T_{50} = T_{49} + (2 \times 49 + 1) = 2401 + 99 = 2500 = 50^2

✓.

5Problem 5

Answer

(a)

n + (n+1) + (n+2) = 96

(b) 31, 32, 33 (c) Sum

= 3n + 3 = 3(n+1)

(d) No — sum

= 4n + 6 = 2(2n + 3)

, divisible by 2 not 4

Full working

(a)

n + (n+1) + (n+2) = 3n + 3 = 96

.

(b)

n = 31

. Integers: 31, 32, 33.

(c) Sum

= 3n + 3 = 3(n + 1)

— always a multiple of 3, in fact

3 \times

middle integer.

(d) Sum of 4 consecutive integers

= 4n + 6 = 2(2n + 3)

. Since

2n + 3

is odd, this is

2 \times \text{odd}

, divisible by 2 but **not by 4**.

6Problem 6

Answer

(a)

T_n = 2n + 1

(b) 40 triangles (c) 0 sticks left over (uses all 81)

Full working

(a) Common difference 2,

T_1 = 3

→

T_n = 2n + 1

.

(b) Solve

2n + 1 \leq 81 \Rightarrow n \leq 40

. Make 40 triangles, using

2 \times 40 + 1 = 81

sticks.

(c) Zero sticks left over.

7Problem 7

Answer

(a) 1, 3, 6, 10, 15 (b)

T_n = n(n+1)/2

(c)

T_{50} = 1275

(d) See working

Full working

(a) Cumulative sums: 1, 3, 6, 10, 15.

(b)

T_n = n(n+1)/2

.

(c)

T_{50} = 50 \times 51 / 2 = 1275

.

(d) Pair:

1 + n

2 + (n-1)

, etc — each pair sums to

n + 1

. There are

n/2

pairs (when

n

even). Sum

= \tfrac{n(n+1)}{2}

. (For odd

n

, the same identity holds via the algebraic argument.)

8Problem 8

Answer

(a) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 (b)

F_{10} = 55

\varphi \approx 1.618

Full working

(a) Each term is the sum of the two before: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55.

(b)

F_{10} = 55

.

(c)

F_6/F_5 = 8/5 = 1.6

F_9/F_8 = 34/21 \approx 1.619

F_{11}/F_{10} = 89/55 \approx 1.618

. As

n \to \infty

, the ratio approaches the **golden ratio**

\varphi = (1 + \sqrt{5})/2 \approx 1.61803

9Problem 9

Answer

(a) 55 (b) 185 (c)

1+3+5+7+9 = 25 = 5^2

✓

Full working

(a)

a = 1, d = 1, n = 10

S_{10} = 10(2 + 9)/2 = 10 \times 11 / 2 = 55

.

(b) Last term 32,

d = 3

n

5 + (n-1)3 = 32 \Rightarrow n = 10

S = 10(5 + 32)/2 = 185

.

(c)

1 + 3 + 5 + 7 + 9 = 25 = 5^2

✓.

10Problem 10

Answer

(a)

\tfrac{1}{3}, \tfrac{35}{99}, \tfrac{17}{999}

(b) Let

x = 0.\overline{9}

, then

10x - x = 9

, so

x = 1

Full working

(a)

\tfrac{3}{9} = \tfrac{1}{3}

;

\tfrac{35}{99}

;

\tfrac{17}{999}

.

(b)

x = 0.\overline{9}

10x = 9.\overline{9}

. Subtract:

9x = 9

x = 1

. ∎

(c) Partial sums:

0.9, 0.9 + 0.09 = 0.99, 0.999, \ldots

Each partial sum gets closer to 1 (difference:

0.1, 0.01, 0.001, \ldots

). The limit is exactly 1, formalising the algebra of part (b).

11Problem 11

Answer

(a) 2, 4, 8, 16, 32 (b)

L_n = 2^n

Full working

(a)

L = 2, 4, 8, 16, 32

.

(b) Geometric,

L_n = 2^n

.

(c)

L_{10} = 1024

.

(d)

L_{12} = 4096

layers.

12Problem 12

Answer

(a) 56, 72 (b) First diffs 4, 6, 8, 10, 12; second diffs 2 (constant) → quadratic (c)

1 \times 2 = 2

2 \times 3 = 6

, etc. (d) 420

Full working

(a) Differences: 4, 6, 8, 10, 12, 14, 16. So next two:

42 + 14 = 56

;

56 + 16 = 72

.

(b) First differences increase by 2 → second difference = 2 (constant). So the sequence is quadratic,

T_n = an^2 + bn + c

with

2a = 2

, i.e.

a = 1

.

(c)

T_n = n(n+1)

T_1 = 2, T_2 = 6, T_3 = 12, T_4 = 20, T_5 = 30

✓.

(d)

T_{20} = 20 \times 21 = 420