Active Maths 2 (Strands 1–5): Ch 4 Solutions 1 Chapter 4 Exercise 4.1 Q. 1. (i) Pattern repeats every 5th tile 12 ÷ 5 = 2 R 2 ∴ The 12th tile is a square (ii) 35 ÷ 5 = 7 R 0 ∴ The 35th tile is a circle (iii) 105 ÷ 5 = 21 R 0 ∴ The 105th tile is a circle (iv) 206 ÷ 5 = 41 R 1 ∴ The 206th tile is a triangle Q. 2. (i) Pattern repeats every 6th marble 22 ÷ 6 = 3 R 4 ∴ The 22nd marble is green (ii) 61 ÷ 6 = 10 R 1 ∴ The 61st marble is red (iii) 27 ÷ 6 = 4 R 3 Full pattern repeats 4 times giving 4 × 2 = 8 red marbles plus 2 more from the next 3 marbles. ∴ A red marble appears 10 times Q. 3. (i) Adding 3 each time ∴ Pattern 4 needs 13 matchsticks (ii) Adding 3 more pattern 5 needs 16 matchsticks (iii) ____________ Pattern 4 ______________ Pattern 5 (iv) Total number of matchsticks needed to make the first 7 patterns = 4 + 7 + 10 + 13 + 16 + 19 + 22 = 91 matchsticks Q. 4. (i) 2, 6, 10, 14 … (a) T 1 (start term) = 2 (b) 14 2 6 10 +4 +4 +4 The common difference is +4 (c) T 5 (5th term): 14 + 4 = 18 T 6 (6th term): 18 + 4 = 22 T 7 (7th term): 22 + 4 = 26 The next 3 terms are 18, 22, 26 (d) Start with 2 and then add 4 to every term.
Transcript
Active Maths 2 (Strands 1–5): Ch 4 Solutions 1
Chapter 4 Exercise 4.1
Q. 1. (i) Pattern repeats every 5th tile
12 ÷ 5 = 2 R 2
∴ The 12th tile is a square
(ii) 35 ÷ 5 = 7 R 0
∴ The 35th tile is a circle
(iii) 105 ÷ 5 = 21 R 0
∴ The 105th tile is a circle
(iv) 206 ÷ 5 = 41 R 1
∴ The 206th tile is a triangle
Q. 2. (i) Pattern repeats every 6th marble
22 ÷ 6 = 3 R 4
∴ The 22nd marble is green
(ii) 61 ÷ 6 = 10 R 1
∴ The 61st marble is red
(iii) 27 ÷ 6 = 4 R 3
Full pattern repeats 4 times giving 4 × 2 = 8 red marbles plus 2 more from the next 3 marbles.
(iv) Total number of matchsticks needed to make the first 7 patterns
= 4 + 7 + 10 + 13 + 16 + 19 + 22
= 91 matchsticks
Q. 4. (i) 2, 6, 10, 14 …
(a) T1 (start term) = 2
(b) 142 6 10
+4+4 +4
The common difference is +4
(c) T5 (5th term): 14 + 4 = 18
T6 (6th term): 18 + 4 = 22
T7 (7th term): 22 + 4 = 26
The next 3 terms are 18, 22, 26
(d) Start with 2 and then add 4 to every term.
2 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii) 7, 19, 31, 43 …
(a) T1 = 7
(b) 437 19 31
+12+12 +12
The common difference is +12
(c) T5: 43 + 12 = 55
T6: 55 + 12 = 67
T7: 67 + 12 = 79
The next 3 terms are 55, 67, 79
(d) Start with 7 and then add 12 to every term.
(iii) 9, 24, 39, 54 …
(a) T1 = 9
(b) 549 24 39
+15+15 +15
The common difference is +15
(c) T5: 54 + 15 = 69
T6: 69 + 15 = 84
T7: 84 + 15 = 99 The next 3 terms are 69, 84, 99
(d) Start with 9 and then add 15 to every term.
(iv) 3, −5, −13, −21 …
(a) T1 = 3
(b) –213 –5 –13
–8–8 –8
The common difference is −8
(c) T5: −21 −8 = −29
T6: −29 − 8 = −37
T7: −37 − 8 = −45
The next 3 terms are −29, −37, −45
(d) Start with 3 and then subtract 8 every term.
(v) 1, −10, −21, −32 …
(a) T1 = 1
(b) –321 –10 –21
–11–11 –11
The common difference is −11
(c) T5: −32 −11 = −43
T6: −43 −11 = −54
T7: −54 −11 = −65
The next 3 terms are −43, −54, −65
(d) Start with 1 and then subtract 11 every term.
3Active Maths 2 (Strands 1–5): Ch 4 Solutions
Q. 5. (i) 6, 9, 12, 15 …
(a) T1 = 6
(b) 156 9 12
+3+3 +3
The common difference is +3
(c) T1: 3 × 1 + 3 = 6
T2: 3 × 2 + 3 = 9
∴ Tn: 3 × n + 3
The nth term (general term): Tn = 3n + 3
(d) T9: 3 × 9 + 3 = 30
The 9th term is 30
(ii) 9, 14, 19, 24 …
(a) T1 = 9
(b) 249 14 19
+5+5 +5
The common difference is +5
(c) T1: 5 × 1 + 4 = 9
T2: 5 × 2 + 4 = 14
∴ Tn: 5 × n + 4 The nth term Tn = 5n + 4
(d) T13: 5 × 13 + 4 = 69
The 13th term is 69
(iii) 0, −3, −6, −9 …
(a) T1 = 0
(b) –90 –3 –6
–3–3 –3
The common difference is −3
(c) T1: −3 × 1 + 3 = 0
T2: −3 × 2 + 3 = −3
∴ Tn: −3 × n + 3 Tn = −3n + 3
(d) T19: −3 × 19 + 3 = −54
The 19th term is −54
(iv) −10, −7, −4, −1 …
(a) T1 = −10
(b) –1–10 –7 –4
+3+3 +3
The common difference is +3
(c) T1: 3 × 1 − 13 = −10
T2: 3 × 2 − 13 = −7
∴ Tn: 3 × n − 13 Tn = 3n − 13
(d) T52: 3 × 52 − 13 = 143
The 52nd term is 143
4 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(v) 1000, 748, 496, 244 …
(a) T1 = 1000
(b) 2441,000 748 496
–252–252 –252
The common difference is −252
(c) T1: −252 × 1 + 1252 = 1,000
T2: −252 × 2 + 1252 = 748
∴ Tn: −252 × n + 1252 Tn = −252n + 1252
(d) T100: −252 × 100 + 1252 = −23,948
The 100th term is −23,948
Q. 6. (i) Tn = 3n
(a) T1: 3 × 1 = 3
(b) T3: 3 × 3 = 9
(c) T45: 3 × 45 = 135
(ii) Tn = 2n + 5
(a) T1: 2 × 1 + 5 = 7
(b) T3: 2 × 3 + 5 = 11
(c) T45: 2 × 45 + 5 = 95
(iii) Tn = n − 3
(a) T1: 1 − 3 = −2
(b) T3: 3 − 3 = 0
(c) T45: 45 − 3 = 42
(iv) Tn = 4 − n
(a) T1: 4 − 1 = 3
(b) T3: 4 − 3 = 1
(c) T45: 4 − 45 = −41
(v) Tn = 9n − 1
(a) T1: 9 × 1 − 1 = 8
(b) T3: 9 × 3 − 1 = 26
(c) T45: 9 × 45 − 1 = 404
(vi) Tn = 3 − 2n
(a) T1: 3 − 2 × 1 = 1
(b) T3: 3 − 2 × 3 = −3
(c) T45: 3 − 2 × 45 = −87
Q. 7. (i) Student A Student B Student C
Start amount €10 €0 €90
End amount €130 €600 €180
Amount saved over 6 wks €120 €600 €90
Amount saved per wk €20 €100 €15
(ii) Using the general term to find the formula for the amount each student saves. Let m = amount of money saved t = time in weeks Student A Common difference (amount saved per week) = (130 − 10) ÷ 6 = 20 Starting amount at zero weeks = 10 ∴ m = 20t + 10 Student B Common difference = (600 − 0) ÷ 6 = 100 Starting amount at zero weeks = 0 ∴ m = 100t + 0 m = 100t
5Active Maths 2 (Strands 1–5): Ch 4 Solutions
Student C
Common difference = (180 − 90) ÷ 6
= 15
Starting amount at zero weeks = 90
∴ m = 15t + 90
(iii) After 12 weeks
Student A Student B Student Cm = 20t + 10 m = 100t m = 15t + 90
if t = 12, m = 20 × 12 + 10 if t = 12, m = 100 × 12 if t = 12, m = 15 × 12 + 90m = 250 m = 1,200 m = 270
Student A has €250 Student B has €1,200 Student C has €270
(iv) The graph for Student B is directly proportional, as the line goes through the origin.
(v) Constant of proportionality = 100
In the context of the question, the constant of proportionality means that for every week that goes by, the amount saved increases by €100.
Q. 8. (i) Pattern 1: 7 matchsticks
Pattern 2: 12 matchsticks
Pattern 3: 17 matchsticks
+5 +5 +5 +5 +5
22 27 327 12 17
Adding 5 each time pattern 6 needs 32 matchsticks.
(ii) T1: 5 × 1 + 2 = 7
∴ Tn: 5 × n + 2 = 5n + 2
Let m = number of matchsticks
n = pattern number
∴ m = 5n + 2
(iii) The first pattern will take 5 + 2 i.e. 7 matches and thereafter there are an additional 5 matches for each new pattern.
(iv) Pattern 25 i.e. n = 25
∴ m = 5 × 25 + 2
= 127
Pattern 25 needs 127 matches
(v) 182 matchsticks used i.e. m = 182
∴ 182 = 5n + 2
180 = 5n
36 = n
Pattern 36 requires 182 matchsticks
6 Active Maths 2 (Strands 1–5): Ch 4 Solutions
Q. 9. (i) Adding 3 extra marbles each time
∴ The 5th pattern needs 18 marbles.
(ii) T1 = 6
Common difference = 3 T1: 3 × 1 + 3 = 6 ∴ Tn: 3 × n + 3 = 3n + 3 The nth pattern needs 3n + 3 marbles. (iii) The number of marbles needed for any pattern in the sequence is 3 times the
pattern number plus an extra 3 marbles (iv) 93 marbles ⇒ 3n + 3 = 93 3n = 90 n = 30 The 30th pattern requires 93 marbles.
Q. 10. (i) 4, 7, 10 …
Adding 3 each time
∴ The 7th pattern needs 22 coins.
(ii) T1: 3 × 1 + 1 = 4
T2: 3 × 2 + 1 = 7
∴ Tn: 3 × n + 1 = 3n + 1
The nth pattern needs 3n + 1 coins.
(iii) 4 + 7 + 10 + 13 + 16 = 50
∴ The student will have made 5 complete patterns.
Q. 11. (i) Time (mins) Cost (€) Company A Cost (€) Company B0 0.3 0.382 0.37 0.434 0.44 0.486 0.51 0.538 0.58 0.58
10 0.65 0.6312 0.72 0.6814 0.79 0.7316 0.86 0.78
(ii) Company A: cost for 2 minutes is €0.44 − €0.37 = €0.07 ∴ Cost for 1 minute = €0.035 (iii) Company B: cost for 2 minutes is €0.48 − €0.43 = €0.05 ∴ Cost for 1 minute = €0.025 (iv) Company B has a higher connection fee of 38 cents (€0.38), which is higher than
company B by 8 cents (€0.08). (v) Company A costs an initial 30 cents (€0.30) to make a phone call plus an additional
3.5 cents (€0.035) per minute. Company B costs an initial 38 cents (€0.38) to make a phone call plus an additional
2.5 cents (€0.025) per minute.
7Active Maths 2 (Strands 1–5): Ch 4 Solutions
(vi) c = Cost (€), m = number of minutes
Company A: c = 0.035 m + 0.3
Company B: c = 0.025 m + 0.38
(vii) From the graph an 8-minute call costs the same from both companies (i.e. €0.58).
(viii) cA = 0.035 m + 0.3
cB = 0.025 m + 0.38
Solve to find when cA = cB i.e. when the cost of calls from companies A and B are the same:
0.035 m + 0.3 = 0.025 m + 0.38
0.01 m = 0.08
m = 0.08 ____ 0.01
m = 8 minutes
(ix) Company A: c = 0.035 m + 0.3
if m = 23 minutes then c = 0.035 × 23 + 0.3
∴ c = 1.105
Company A charges €1.105 for a 23-minute call.
Company B: c = 0.025 m + 0.38
if m = 23 minutes then c = 0.025 × 23 + 0.38
c = 0.955
Company B changes €0.955 for a 23-minute call.
Q. 12. (i) Company A: from the graph, 10 km costs €50.
∴ 1 km costs €5
∴ Company A charges €5/km.
Company B: from the graph, 50 km costs €150 [i.e. €200 − €50 = €150]
∴ 1 km costs €3
∴ Company B charges €3/km
(ii) Company B has a standing charge of €50.
(iii) The trip would cost €5 per km with company A.
The trip would cost €50 plus €3 per km with company B.
(iv) c = cost (€), d = kilometres travelled
Company A Company Bcommon difference = 5 common difference = 3
standing charge = 0 standing charge = 50
∴ c = 5d + 0 ∴ c = 3d + 50
c = 5d
(v) From the graph at about 25 km both companies charge the same price.
8 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(vi) When both companies charge the same price
5d = 3d + 50
2d = 50
∴ d = 25. Distance equals 25 km as required.
(vii) Budget of €500
Company A Company Bc = 5d c = 3d + 50
∴ 500 = 5d 500 = 3d + 50
100 = d 450 = 3d
150 = d
The teacher will use Company B, as this gives the greatest distance.
The total distance covered on the tour is 150 km.
(viii) Company A, because there is no standing charge; charges are only incurred per kilometre travelled.
(iv) The formula means that you start with 50 (litres) of water and subtract 2.5 (litres) every hour. The subtraction describes the leak from the water storage tank.
(v) After 6 hours
V = −2.5t + 50
∴ V = −2.5 × 6 + 50
V = 35 litres
(vi) Time for tank to empty
if V = 0 then −2.5t + 50 = 0
50 = 2.5t
20 = t
It takes 20 hours for the tank to empty.
Q. 14. (i)
0 1 2 3 4 5
Tonne (metal)
6 7 8
x
y
0
5
10
No
. o
f cars
15
20
25
9Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii) 8 tonnes = 20 cars
1 tonne = 2.5 cars
Slope of line = 2.5
∴ Constant of proportionality = 2.5
(iii) For every tonne of metal, 2.5 cars are produced.
(iv) m = 2.5c
(v) 2.5c = m
c = m ___ 2.5 or 0.4 m
∴ c = 0.4 m
(vi) c = 0.4(525) = 210 cars
(vii) m = 2.5(125) = 312.5 tonnes of metal
Q. 15. (i) Time (hours)
Electrician A
Electrician B
0 (call-out) 50 0
1 90 60
2 130 120
3 170 180
4 210 240
5 250 300
(ii) Graph B is directly proportional, as it passes through the origin and the equation of this line would be in the form y = mx, where m the slope is 60.
= 1 + 2 + 3 + 4 + … + 25 + 26 = 351 letters 1,000 ÷ 351 = 2 R 298 Working backwards from 351 − 26 (letter Z) = 325 325 − 25 (letter Y) = 300 ∴ The 1,000th letter is X
Q. 19.
–5 –5 –5 –5
64 5979 74 69 and
+4 +4 +4 +4
19 237 11 15
T1 = −5 × 1 + 84 T1 = 4 × 1 + 3
Tn = −5n + 84 Tn = 4n + 3
If the mth sequences are the same then
−5 m + 84 = 4 m + 3
81 = 9 m
9 = m
The 9th term is the same.
12 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(b) The value of the 9th term is
T9 = −5 × 9 + 84 or T9 = 4 × 9 + 3
T9 = 39 T9 = 39
The value of the 9th term is 39
Q. 20. (i) Area (m2) Cost (€)
1 75
2 150
3 225
4 300
5 375
6 450
7 525
(ii)
00
50
100
150
200
250
300
350
400
450
500
550
600
1 2 3 4 5 6 7
Area (m2)
Co
st
(€)
x
y
(iii) Slope of line = 75
∴ Constant of proportionality = 75
(iv) Cost of shed (C)
Area per square metre (A)
C = 75A
(v) C = 75(12) = €900
Q. 21. (i) 6 blue counters in pattern 6
(ii) 22 red counters in pattern 7
(iii) 5, 9, 13, 17, 21, 25, 29, 33, 37
Adding 4 each time, pattern 9 needs 37 counters.
(iv) T1 = 4 × 1 + 1
T2 = 4 × 2 + 1
Tn = 4n + 1
∴ There are 4n + 1 counters in the nth pattern.
(vi) C = 75A
∴ A = C ___ 75
A = 4,500
______ 75 = 60 m2
(vii) €1,170 = 18 m2
€65 = 1 m2
(viii) Company A Company B
1 m2 = €75 1 m2 = €65
30 m2 = €2,250 30 m2 = €1,950
Company B charges €300 less than Company A to build a shed of 30 m2.
13Active Maths 2 (Strands 1–5): Ch 4 Solutions
(v) Red counters in pattern
+3 +3
4 7 10 …
common difference = +3 T1 = 3 × 1 + 1 Tn = 3n + 1 ∴ There are 3n + 1 red counters in the nth pattern. (vi) Blue counters in pattern 1, 2, 3 … common difference = +1 T1 = 1 × 1 T2 = 1 × 2 ∴ Tn = 1 × n Tn = n ∴ There are n counters in the nth pattern. (vii) (a) The number of red counters in the pattern is calculated by 3 times the pattern
number plus 1 (b) The number of blue counters in the pattern is the same as the pattern number.
Q. 22. (i) Blue tiles
+2 +2
1 3 5… common difference = +2
There will be 21 blue tiles in pattern 11.
(ii) Tiles in pattern
+3 +3
1 4 7… common difference = +3
There will be 58 tiles in pattern 20.
(iii) (a) Red tiles
+1 +1
0 1 2 …
T1 = 1 × 1 − 1 T2 = 1 × 2 − 1 ∴ Tn = 1 × n − 1 Tn = n − 1 (b) Blue tiles
(ii) The relationship between dollars and euros is directly proportional because we can see from the graph that the line passes through (0,0), i.e. 0 dollar converts to 0 euros.
(iii) y-axis represents dollars = d
x-axis represents euros = e
∴ d = 1.25e
(iv) Constant of proportionality = 1.25
(v) The constant of proportionality represents the exchange rate, i.e. €1 = $1.25.
(vi) Henry will receive $1,000
(vii) Formula: d = 1.25e
d = 1.25 (800)
= 1,000
Q. 24. (i)
0 1 2 3 4 5
50
100
150
200
Ch
arg
es
(€
)
Mechanic A
Mechanic B
Time (hours)
15Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii) The call-out fee for mechanic A is €75.
(iii) Let c = mechanic’s charge (€) and t = time (hours)
Mechanic A Mechanic Bcommon difference = (200 − 150) ÷ 2 common difference = (200 − 100) ÷ 2
(c) After 5 seconds both runners are the same distance from Tina.
(d) After 9 seconds Jenny is furthest from Tina and 4 m ahead of Bill
(e) The distance between Bill and Tina increases by 2 m each second. ∴ Common difference = +2
Let d = distance between runners t = time in seconds Starting distance = 7 (when t = 0) ∴ d = 2 × t + 7 d = 2t + 7
(f) The distance between Jenny and Tina increases by 3 m every second.
∴ Common difference = +3 Starting distance = 2 (when t = 0) ∴ d = 3 × t + 2 d = 3t + 2
(g) Using formulas in (e) and (f) to verify the answer to part (c) that both runners are the same distance from Tina after 5 seconds.
3t + 2 = 2t + 7 t = 5 as required
17Active Maths 2 (Strands 1–5): Ch 4 Solutions
(h) Jenny stops after 1 min (60 seconds) Jenny: d = 3 × 60 + 2 d = 182
after 1 minute Jenny is 182 m from Tina.
Bill: d = 2 × 60 + 7
d = 127
after 1 minute Bill is 127 m from Tina.
Distance for Bill to catch up = 182 − 127 = 55 m
At 2 m/s it will take (55 ÷ 2) i.e. 27 1 __ 2 seconds
for Bill to be level again.
(i) –2–5 –4 –3 –1 0 1 2 ...
+1+1 +1
The distance between the two runners changes by 1 m every second.
T0: 1 × 0 − 5 = −5
T1: 1 × 1 − 5 = −4
∴ Tn: 1 × n − 4 = n − 5
Letting d = distance between runners and t = time in seconds gives d = t − 5
thus when d = 100
100 = t − 5
105 = t
Runners would be 100 m apart after 105 seconds
Q. 27. (a) From the data in the table there is a common difference of €18 for every 100 units used. This shows that the relationship between the number of units used and the cost is linear.
e.g. 56 − 38 = 18
74 − 56 = 18
92 − 74 = 18 etc.
(b)
100 200 300 400 500 600 700 800
20
40
60
80
100
120
140
160
Co
st
(€)
Units used
Plan A
Plan B
18 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(c) From the graph the standing charge is approximately €20.
(d) Common difference = +18 per 100 units
T1: 18 × 1 + 20 = 38
T2: 18 × 2 + 20 = 56
∴ Standing charge when no units used is €20
(e) C = 0.18u + 20, where C is the cost in euro and u is the units used
(f) C = 0.18(650) + 20
= 137
155.50 − 137 = 18.50
Vat rate% = 18.50 ______ 137 × 100
= 13.5% to one decimal place (g) Units used 100 200 300 400 500 600 700 800
Plan B cost (€) 51.50 67.00 82.50 98.00 113.50 129.00 144.50 160.00
Standing charge = €36, Rate per unit = 15.5 cent
Rate per 100 units = €15.50
(h) If Lisa used 650 units on plan B that would cost 36 + 0.155 × 650 = €136.75, which is only 25 cents less than Plan A. Plan B only works out to be cheaper for
higher users of electricity.
(i)
1020
4030
6070
50
8090
100110120130140150160170
100 200 300 400 500 600 700 800
Co
st
(€)
Units used
Plan A
Plan B
(j) From the graph, both plans are equal at approximately 640 units.
Algebraically, 18n + 20 = 15.5n + 36
2.5n = 16
n = 6.4 n = hundreds of units
∴ Number of units = 640
19Active Maths 2 (Strands 1–5): Ch 4 Solutions
Exercise 4.2
Q. 1. (i) 14, 15, 21, 32 …
(a) T1 (start term) = 14
(b) 3214 15 21
+11+1 +6
+5 +5
The first differences: 1, 6, 11 …
The second difference: +5
(c) T4: 21 + 11 = 32
∴ T5: 32 + 16 = 48
T6: 48 + 21 = 69
T7: 69 + 26 = 95
The next 3 terms are 48, 69, 95
(ii) 16, 26, 42, 64 …
(a) T1 = 16
(b) 6416 26 42
+22+10 +16
+6 +6
The first differences: 10, 16, 22
The second difference: +6
(c) T4: 42 + 22 = 64
T5: 64 + 28 = 92
T6: 92 + 34 = 126
T7: 126 + 40 = 166
The next 3 terms are 92, 126, 166
(iii) 0, 4, 11, 21 …
(a) T1 = 0
(b) 210 4 11
+10+4 +7
+3 +3
The first differences: 4, 7, 10
The second difference: +3
(c) T4: 11 + 10 = 21
T5: 21 + 13 = 34
T6: 34 + 16 = 50
T7: 50 + 19 = 69
The next 3 terms are 34, 50, 69
(iv) 5, 7, 6, 2
(a) T1 = 5
(b) 25 7 6
–4+2 –1
–3 –3
The first differences: 2, −1, −4
The second difference: −3
(c) T4: 6 − 4 = 2
T5: 2 − 7 = –5
T6: −5 − 10 = −15
T7: −15 − 13 = −28
The next 3 terms are −5, −15, −28
(v) −8, −12, −10, −2 …
(a) T1 = −8
(b) –2–8 –12 –10
+8–4 +2
+6 +6
The first differences: −4, 2, 8
The second difference: +6
(c) T4: −10 + 8 = −2
T5: −2 + 14 = 12
T6: 12 + 20 = 32
T7: 32 + 26 = 58
The next 3 terms are 12, 32, 58
Q. 2. (i) 3, 6, 12, 24 …
(a) T1: 3
T2: 3 × 2 = 6
T3: 6 × 2 = 12
T4: 12 × 2 = 24
The sequence doubles.
(b) T5: 24 × 2 = 48
T6: 48 × 2 = 96
20 Active Maths 2 (Strands 1–5): Ch 4 Solutions
T7: 96 × 2 = 192
The next 3 terms : 48, 96, 192
(ii) 13, 39, 117, 351 …
(a) T1: 13
T2: 13 × 3 = 39
T3: 39 × 3 = 117
The sequence triples.
(b) T5: 351 × 3 = 1053
T6: 1053 × 3 = 3159
T7: 3159 × 3 = 9477
The next 3 terms are 1053, 3159, 9477
(iii) 1 __ 8 , 1 __ 4 , 1 __ 2 , 1 …
(a) T1: 1 __ 8
T2: 1 __ 8 × 2 __ 1 = 1 __ 4
T3: 1 __ 4 × 2 __ 1 = 1 __ 2
The sequence doubles.
(b) T5: 1 × 2 = 2
T6: 2 × 2 = 4
T7: 4 × 2 = 8
The next 3 terms: 2, 4, 8
(iv) 1 ___ 81 , 1 ___ 27 , 1 __ 9 , 1 __ 3 …
(a) T1: 1 ___ 81
T2: 1 ___ 81 × 3 __ 1 = 1 ___ 27
T3: 1 ___ 27 × 3 __ 1 = 1 __ 9
∴ The sequence triples.
(b) T5: 1 __ 3 × 3 __ 1 = 1
T6: 1 × 3 __ 1 = 3
T7: 3 × 3 = 9
The next 3 terms: 1, 3, 9
(v) −2.5, −7.5, −22.5, −67.5 …
(a) T1: −2.5
T2: –2.5 × 3 = –7.5
T3: −7.5 × 3 = −22.5
The sequence triples.
(b) T4: −22.5 × 3 = −67.5
T5: −67.5 × 3 = −202.5
T6: −202.5 × 3 = −607.5
T7: −607.5 × 3 = −1,822.5
The next 3 terms are −202.5, −607.5, −1,822.5
Q. 3. (i) 37 ...4 11 22
+15+7 +11
+4 +4
The sequence is quadratic, as the second difference is constant.
(ii) 32 ...23 26 29
+3+3 +3
The sequence is linear, as the first difference is constant.
(iii) 161 3 8
+8+2 +5
+3 +3
The sequence is quadratic, as the second difference is constant.
(iv) 2, 4, 8, 16 …
T1: 2
T2: 2 × 2 = 4
T3: 4 × 2 = 8
T4: 8 × 2 = 16
The sequence is exponential, as each term is double the previous term.
(v) –2–11 –8 –5
+3+3 +3
The sequence is linear, as the first difference is constant.
Q. 4. 1, 2, 4 …
(i) Exponential doubling sequence
∴ 4 × 2 = 8 blocks needed for diagram 4.
21Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii) Diagram 5 needs 8 × 2 = 16 blocks
(iii) Diagram 6: 16 × 2 = 32
Diagram 7: 32 × 2 = 64
Diagram 8: 64 × 2 = 128
Diagram 9: 128 × 2 = 256
Diagram 10: 256 × 2 = 512
Diagram 10 needs 512 blocks
(iv) If the pattern was quadratic
1 2 4 ...
+1 +2
+1
T3: 2 + 2 = 4
T4: 4 + 3 = 7
∴ Diagram 4 would need 7 blocks.
(v) T5: 7 + 4 = 11
Diagram 5 would need 11 blocks.
Q. 5. (i) Tn = n2 + 1
(a) T1: 12 + 1 = 2
(b) T2: 22 + 1 = 5
(c) T3: 32 + 1 = 10
Square the number and add 1.
(ii) Tn = 2n2 + n − 2
(a) T1: 2 × 12 + 1 − 2 = 1
(b) T2: 2 × 22 + 2 − 2 = 8
(c) T3: 2 × 32 + 3 − 2 = 19
Square the number, then double it, then add the original number and subtract 2.
(iii) Tn = −n2 − 3n
(a) T1: −(1)2 − 3 × 1 = –4
(b) T2: −(2)2 − 3 × 2 = −10
(c) T3: −(3)2 − 3 × 3 = −18
Square the number, then make it negative, then subtract 3 times the original number.
(iv) Tn = −n2 − 4n + 1
(a) T1: −(1)2 − 4 × 1 + 1 = −4
(b) T2: −(2)2 − 4 × 2 + 1 = −11
(c) T3: −(3)2 − 4 × 3 + 1 = −20
Square the number, then make it negative, then subtract 4 times the original number and add 1.
(v) Tn = 3n
(a) T1: 31 = 3
(b) T2: 32 = 9
(c) T3: 32 = 27
Three to the power of the nth number
Q. 6. (i) 39 ...18 21 28
+11+3 +7
+4 +4
2a = 4
∴ a = 2
Tn = an2 + bn + c
T1: 2 × 12 + b × 1 + c = 18
b + c = 16
T2: 2 × 22 + b × 2 + c = 21
2b + c = 13
Solving simultaneously:
2b + c = 13
− (b + c = 16)
b = −3
Using b + c = 16
gives −3 + c = 16
∴ c = 19
Since a = 2, b = −3 and c = 19
⇒ Tn = 2n2 − 3n + 19
(ii) 32 ...2 7 17
+15+5 +10
+5 +5
2nd difference constant ∴ Sequence is quadratic.
Tn = an2 + bn + c
c
c
c
c
22 Active Maths 2 (Strands 1–5): Ch 4 Solutions
Second difference = 2a
∴ 2a = 5
a = 2.5
T1 = 2 and a = 2.5
i.e. T1 : 2.5(1)2 + b(1) + c = 2
2.5 + b + c = 2
∴ b + c = −0.5
Similarly T2 = 7
∴ T2: 2.5(2)2 + b(2) + c = 7
10 + 2b + c = 7
2b + c = −3
Solve the simultaneous equations:
2b + c = −3
− (b + c = −0.5)
b = −2.5
Using b + c = −0.5
⇒ −2.5 + c = −0.5
c = 2
In summary a = 2.5, b = −2.5 and c = 2
Tn = 2.5n2 − 2.5n + 2
(iii) 27...0 2 11
+16+2 +9
+7 +7
2nd difference = 7
∴ 2a = 7
giving a = 3.5
Tn = an2 + bn + c
T1: 3.5(1)2 + b(1) + c = 0
3.5 + b + c = 0
b + c = −3.5
T2: 3.5(2)2 + b(2) + c = 2
2b + c = −12
Solving these simultaneous equations:
2b + c = −12
− (b + c = −3.5)
b = −8.5
Using b + c = −3.5
−8.5 + c = −3.5
c = 5
In summary a = 3.5, b = −8.5 and c = 5 gives
Tn = 3.5n2 − 8.5n + 5
(iv) –2 ...1 –3 –4
+2–4 –1
+3 +3
The 2nd difference = +3
∴ 2a = 3
and a = 1.5
Tn = an2 + bn + c
T1: 1.5(1)2 + b(1) + c = 1
b + c = −0.5
T2: 1.5(2)2 + b(2) + c = −3
2b + c = −9
Solving these simultaneous equations:
2b + c = −9
− (b + c = −0.5)
b = −8.5
using b + c = −0.5
−8.5 + c = −0.5
c = 8
In summary a = 1.5, b = −8.5 and c = 8 gives
Tn = 1.5n2 − 8.5n + 8
(v) –25 ...–10 –9 –14
–11+1 –5
–6 –6
∴ 2a = −6
a = −3
Tn = an2 + bn + c
T1: −3(1)2 + b(1) + c = −10
b + c = −7
T2: −3(2)2 + b(2) + c = −9
2b + c = 3
c
c
c
c
c
c
c
c
c
c
c
c
23Active Maths 2 (Strands 1–5): Ch 4 Solutions
Solving simultaneous equations:
2b + c = 3
− (b + c = −7)
b = 10
b + c = −7 and b = 10
10 + c = −7
∴ c = −17
In summary a = −3, b = 10 and c = −17 gives
Tn = −3n2 + 10n − 17
Q. 7. (i) Pattern Number of blocks1 22 63 12
(ii)
(iii) 20 ...2 6 12
+8+4 +6
+2 +2
2a = 2
∴ a = 1
Tn = an2 + bn + c
∴ T1: 1(1)2 + b(1) + c = 2
b + c = 1
and T2: 1(2)2 + b(2) + c = 6
2b + c = 2
Solving simultaneous equations:
2b + c = 2
− (b + c = 1)
b = 1
using b + c = 1
1 + c = 1
c = 0
∴ Tn = 1 × n2 + 1 × n + 0
Tn = n2 + n
There are n2 + n blocks in the nth term.
(iv) T10: 102 + 10 = 110
(v) The number of blocks needed for the nth pattern is obtained by squaring the number and then adding the number to this.
Since n is a positive number the biggest pattern in the sequence is pattern 9.
Q. 9. (i) Pattern 4
(ii) 1, 5, 13, 25
1st 4, 8, 12
2nd 4, 4
This is a quadratic sequence starting at 1, with a first difference increasing at a linear rate and the second difference increasing at a constant rate of 4.
(iii) T7 = 85 red discs
(iv) 25 ...1 5 13
+12+4 +8
+4 +4
2nd difference constant means the sequence is a quadratic; also 2a = 4
∴ a = 2
Tn = an2 + bn + c
T1: 2(1)2 + b(1) + c = 1
b + c = −1
T2: 2(2)2 + b(2) + c = 5
2b + c = −3
Solving simultaneous equations:
2b + c = −3
− (b + c = −1)
b = −2
Using b + c = −1
−2 + c = −1
c = 1
∴ Tn = 2 × n2 − 2 × n + 1
Tn = 2n2 − 2n + 1
(v) T20: 2(20)2 − 2(20) + 1
= 761
There are 761 discs in the 20th pattern.
Q. 10. (i) 2020 30 30
–10+10 +0
–10 –10
This is a quadratic sequence, as the 2nd difference is constant.
2a = −10
∴ a = −5
Tn = an2 + bn + c
T1: −5(1)2 + b(1) + c = 20
b + c = 25
T2: −5(2)2 + b(2) + c = 30
2b + c = 50
Solving simultaneous equations:
2b + c = 50
− (b + c = 25)
b = 25
Using b + c = 25
25 + c = 25
c = 0
∴ Tn = −5 × n2 + 25 × n + 0
Tn = −5n2 + 25n
Assuming h = height
and t = time
gives h = −5t2 + 25t
c
c
c
c
c
c
c
c
25Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii)
10 2 3 4 x
y
10
20
30
Time (seconds)H
eig
ht
(m)
(iii) After 3.5 seconds
h = −5(3.5)2 + 25(3.5)
h = 26.25
Height of the rocket after 3.5 seconds is 26.25 m
The height from the graph is close at approximately 25 m and will depend on the scale for further accuracy to be achieved.
(iv) Solve for h = 0 (Rocket at ground level)
∴ −5t2 + 25t = 0
−5t(t − 5) = 0
∴ t = 0 and t = 5
The rocket was at ground level at zero seconds and 5 seconds.
Q. 11. (i)
210 3 4 5
4540353025201510
50
Are
a c
ov
ere
d (
m2)
Days
Area covered
(ii) The data follows an exponential pattern, as it is tripling the previous terms.
e.g. T1: 1.5
T2: 1.5 × 3 = 4.5
T3: 4.5 × 3 = 13.5 etc.
(iii) After 11 days
T4 : 13.5 × 3 = 40.5 keep multiplying by 3, 7 more times to get T11
T11 = 40.5 × 3 × 3 × 3 × 3 × 3 × 3 × 3
∴ T11 = 88,573.5
After 11 days 88,573.5 m2 will be covered.
Q.12. (i) 126 10 12
+0+4 +2
–2 –2
As the second difference is constant, this implies a quadratic sequence.
26 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii)
210 3 4 5
141210
86420
Nu
mb
er (
′000)
Time (weeks)
Number (thousands)
(iii) From the graph the maximum number of seagulls is just over 12,000; approximately 12,200.
(iv) 2a = −2, a = −1 and Tn = an2 + bn + c
T1: −1(1)2 + b(1) + c = 6
b + c = 7
T2: −1(2)2 + b(2) + c = 10
2b + c = 14
Solving these equations simultaneously:
2b + c = 14
− (b + c = 7)
b = 7
Using b + c = 7 and b = 7
gives c = 0
In summary a = −1, b = 7 and c = 0 then
Tn: −1 × n2 + 7 × n + 0 = −n2 + 7n
The general term Tn = −n2 + 7n
(v) Assuming b = no. of seagulls and t = time in weeks.
b = −t2 + 7t
If no seagulls then b = 0 and −t2 + 7t = 0
Solving −t2 + 7t = 0
−t(t − 7) = 0
∴ t = 0 OR t = 7
At 0 weeks and 7 weeks, there will be no seagulls left in the colony.
c
cc
c
c
Q. 13. (i)
210 3 4 5
14161820
1210
86420
Sp
eed
(m
/s)
Time (seconds)
Speed
27Active Maths 2 (Strands 1–5): Ch 4 Solutions
(ii) 1711 17 19
–2+6 +2
–4 –4
As the 2nd difference is constant, the graph is a quadratic.
(Using the graph, symmetry between 2 and 4 seconds also implies a quadratic.)
(iii) 2a = −4 ∴ a = −2 and Tn = an2 + bn + c
T1: −2(1)2 + b(1) + c = 11
b + c = 13
T2: −2(2)2 + b(2) + c = 17
2b + c = 25
Solving equations simultaneously:
2b + c = 25
− (b + c = 13)
b = 12
Using b + c = 13 and b = 12 gives
c = 1
In summary a = −2, b = 12 and c = 1 gives
Tn = −2n2 + 12n + 1
Since v = speed of car and t = any given time
then v = −2t2 + 12t + 1
(iv) From the graph, the speed of the car is approximately 18.5 m/s.
Using the formula when t = 2.5
v = −2(2.5)2 + 12(2.5) + 1
v = 18.5 m/s as required
(v) When v = 1 m/s then −2t2 + 12t + 1 = 1
giving −2t2 + 12t = 0
−2t(t − 6) = 0
∴ t = 0 OR t = 6
The speed of the car is 1 m/s at 0 seconds and 6 seconds.
(iii) From the graph approx 22–23 amoebas will be present after 4.5 hours
28 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(iv) N = 2t
When t = 4.5 N = 24.5
N = 22.6 to 1 d.p. (v) 2t = 65,536 216 = 65,536 ∴ After 16 hours
Guess and check can be used to find the value of t, e.g.
210 = 1024 too low
220 = 1,048,576 too high
Q. 15. (i) Pattern Number of green tiles Number of red tiles Total number of tiles1 8 1 92 12 4 163 16 9 25
(ii) Green tiles increasing by +4 ∴ Pattern 6 needs 28 green tiles. (iii) Red tiles increasing by squaring the pattern number 12, 22, 32
∴ Pattern 7 needs 72 red tiles, i.e. Pattern 7 needs 49 red tiles. (iv) Pattern 8 needs 100 tiles in total. (v) Green tiles:
+4 +4
8 12 16 ...
This is a linear pattern, as the 1st difference is constant. T1: 4 × 1 + 4 = 8 T2: 4 × 2 + 4 = 12 ∴ Tn = 4n + 4 There are 4n + 4 tiles in the nth green pattern. (vi) Red tiles: 16 ...1 4 9
+7+3 +5
+2 +2
This is a quadratic pattern, as the second difference is constant.
2a = 2 ∴ a = 1 and Tn = an2 + bn + c T1: 1(1)2 + b(1) + c = 1 b + c = 0 T2: 1(2)2 + b(2) + c = 4 2b + c = 0 Solving simultaneous equations: 2b + c = 0 − (b + c = 0) b = 0 Using b + c = 0 and b = 0 ⇒ c = 0 also. ∴ Tn = 1(n)2 + 0(n) + 0 Tn = n2
There are n2 tiles in the nth red pattern.
29Active Maths 2 (Strands 1–5): Ch 4 Solutions
(vii) 36 ...9 16 25
+11+7 +9
+2 +2
This is a quadratic pattern, as the second difference is constant.
2a = 2 ∴ a = 1 Tn = an2 + bn + c T1: 1(1)2 + b(1) + c = 9 T2: 1(2)2 + b(2) + c = 16 b + c = 8 2b + c = 12 Solving simultaneous equations: 2b + c = 12 − (b +c = 8) b = 4 Using b + c = 8 and b = 4 gives 4 + c = 8 ∴ c = 4 In summary a = 1, b = 4 and c = 4 gives Tn = 1(n)2 + 4(n) + 4 Tn = n2 + 4n + 4 This is a perfect square and can be written also as (n + 2)2. There are n2 + 4n + 4 OR (n + 2)2 tiles in the nth pattern.
Q. 16. (i) Blue tiles 20 ...2 6 12
+8+4 +6
+2 +2
This is a quadratic pattern, as the 2nd difference is constant.
2a = 2 ⇒ a = 1 and Tn = an2 + bn + c T1: 1(1)2 + b(1) + c = 2 T2: 1(2)2 + b(2) + c = 6 ∴ b + c = 1 2b + c = 2 Solving simultaneous equations: 2b + c = 2 − (b + c = 1) b = 1 Since b + c = 1 and b = 1 this gives c = 0 In summary a = 1, b = 1 and c = 0 ∴ Tn: 1(n)2 + 1(n) + 0 = n2 + n Tn = n2 + n (ii) When n = 10 number of blue tiles T10: 102 + 10 = 110 ∴ 110 blue tiles needed (iii) Red tiles 5 ...2 3 4
+1+1 +1
This is a linear pattern, as the 1st difference is constant.
T1: 1 × 1 + 1 = 2
T2: 1 × 2 + 1 = 3
∴ Tn: 1 × n + 1
Tn = n + 1
c
c
c
c
30 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(iv) When n = 17, T17: 17 + 1 = 18 ∴ 18 red tiles needed. (v) Total number of tiles = total blue + total red ∴ Tn = n2 + n + n + 1 Tn = n2 + 2n + 1 OR (n + 1)2
(vi) 650 tiles in total 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 = 649 ∴ 11 patterns in total can be made.
Q. 17. Graph A is a linear pattern, as the slope of the line equals 2, that is the 1st difference is constant.
Graph B is an exponential pattern, as the pattern is doubling, for example, when x = 1, y = 2.
when x = 2, y = 4 when x = 3, y = 8 and when x = 4, y = 16 Graph C is a quadratic pattern because of the symmetry in the shape of the graph
either side of the minimum at x = 4. Looking at the change in the y-values for every 1 unit change in the x-value, the second difference is constant (+2), indicating a quadratic pattern.
116 9 4
–3–7 –5
0 ...
–1
+2 +2 +2
Q. 18. (i) 03 0 –1
+1–3 –1
3
+3
+2 +2 +2
Yes, this graph represents a quadratic pattern, as the second difference is constant. (ii) Between 2 and 4 seconds (below x-axis). (iii) 2nd difference = 2
∴ a = 1 Tn = an2 + bn + c T1 = 3 T2 = 0 1(1)2 + b(1) + c = 3 1(2)2 + b(2) + c = 0 1 + b + c = 3 4 + 2b + c = 0 b + c = 2 2b + c = −4 Solving simultaneous equations: b + c = 2 −(2b + c = −4) −b = 6 b = −6
31Active Maths 2 (Strands 1–5): Ch 4 Solutions
−6 + c = 2
c = 8
∴ Tn = 1n2 − 6n + 8
h = t2 − 6t + 8
(iv) The aeroplane was released at time t = 0.
h = 0 − 6(0) + 8
h = 8 m
(v) h = (20)2 − 6(20) + 8
h = 288 m
(vi) Barry has assumed that the paper aeroplane will continue to fly in a quadratic pattern.
(vii) No, as the paper aeroplane cannot possibly fly that high when launched from 8 m.
Up until the 9th month the shares had been increasing in value; however, in the 10th month they start to fall, i.e. start to lose money. That is why the broker advised that the shares are sold after the 9th month when they reached their peak.
Q. 22. (i) 80 ...5 20 45
+35+15 +25
+10 +10
Yes. The data represents a quadratic pattern, as the 2nd difference of +10 is constant.
(ii) After 4 seconds the coin has fallen 80 m.
(iii) f = distance coin falls in m, t = time in seconds
2a = 10 ∴ a = 5 and Tn = an2 + bn + c
T1: 5(1)2 + b(1) + c = 5 T2: 5(2)2 + b(2) + c = 20
b + c = 0 2b + c = 0
Solving simultaneous equations:
2b + c = 0
−(b + c = 0)
b = 0 ∴ c = 0 also
∴ Tn: 5n2 + 0 + 0 = 5n2
and f = 5t2
(iv) If f = 200 metres then 5t2 = 200
t2 = 40
t = √___
40
t = 6.3 seconds to 1 d.p.
c
cc
c
c
34 Active Maths 2 (Strands 1–5): Ch 4 Solutions
(v) Time (s) Height above ground (m)0 2000.5 198.751 195 [= 200 – 5]1.5 188.75 [= 200 – 11.25]
fall = 5 × time2
5(0.5)2 = 1.25
5(1)2 = 5
5(1.5)2 = 11.25
Subtract the amount fallen at each time interval to find the height.
(vi) h = height of coin above ground (in m)
t = time in seconds
∴ h = 200 − 5t2 OR h = −5t2 + 200
(vii) When h = 50 m, calculate t:
50 = 200 − 5t2
5t2 = 150
t2 = 30
t = √___
30 , t = 5.4772...
At 5.5 seconds (to 1 d.p.)
(viii) If t = 20 seconds, how far did the coin fall?
using f = 5t2
f = 5 × 202
f = 2,000
The coin would have been dropped 2,000 m.
(ix) If h = 10 km (10,000 m)
5t2 = 10,000
t2 = 2,000
t = √______
2,000
t = 44.721…
The coin would have taken 45 seconds (to the nearest second).
