QCAA General Mathematics Growth and decay in sequences

15 sample questions with marking guides and sample answers

Q3
2023
SCSA
Paper 1
9 marks
Q3

From January 1, 2020, a company offered its employees an income package with a starting wage of $4000 per month, paid at the end of each month. Also, as an incentive to stay with the company, there was a monthly increase of $50 each month.

Q3a
2 marks

Determine a recursive rule for the monthly wage.

Reveal Answer

Tn+1=Tn+50,T1=4000T_{n+1} = T_n + 50, T_1 = 4000

Marking Criteria
DescriptorMarks

states correct recursive rule

1

states correct first term

1
Q3b
2 marks

Deduce a simplified rule for the nnth term of the monthly wage.

Reveal Answer

Tn=4000+(n1)(50)T_n = 4000 + (n-1)(50)
Tn=3950+50nT_n = 3950 + 50n

Marking Criteria
DescriptorMarks

uses correct arithmetic formula

1

gives correct simplified rule for the nthn^{\text{th}} term

1
Q3c
2 marks

Determine the monthly wage for December 2020.

Reveal Answer

T12=3950+50(12)=4550T_{12} = 3950 + 50(12) = 4550

Therefore, the monthly wage for December 2020 is $4550

Marking Criteria
DescriptorMarks

correctly identifies term 12

1

correctly calculates the $4550

1
Q3d
3 marks

The company has decided to make the monthly increase $60 from the end of December 2023.

Calculate the monthly wage for March 2024.

Reveal Answer

12×4=4812 \times 4 = 48
T48=3950+50(48)=6350T_{48} = 3950 + 50(48) = 6350

Therefore, the monthly wage for March 2024 is 6350+60+60+60=65306350 + 60 + 60 + 60 = $6530

Marking Criteria
DescriptorMarks

correctly calculates T48T_{48}

1

calculates correct term for March 2024

1

states correct solution for wage

1
Q8
2024
QCAA
Paper 1
1 mark
Q8
1 mark

After nn bounces, the rebound height (cm) of a ball, tnt_n, is modelled by the rule tn=240×0.5(n1)t_n = 240 \times 0.5^{(n-1)}.
Calculate the difference in rebound height (cm) between the first bounce and the third bounce.

A

90

B

120

C

180

D

210

Reveal Answer
A

90

This value results from an incorrect calculation of the bounce heights or their difference.

B

120

This is the height of the second bounce (t2=120t_2 = 120) or the difference between the first and second bounces, rather than the difference between the first and third.

C

180

Correct Answer

First calculate t1=240×0.50=240t_1 = 240 \times 0.5^0 = 240 and t3=240×0.52=60t_3 = 240 \times 0.5^2 = 60. The difference is 24060=180240 - 60 = 180.

D

210

This error likely occurs if the formula 240×0.5n240 \times 0.5^n is used instead of 240×0.5(n1)240 \times 0.5^{(n-1)}, yielding a third bounce of 30 and a difference of 24030=210240 - 30 = 210.

Q2
2025
QCAA
Paper 1
1 mark
Q2
1 mark

Which option shows three consecutive terms for an arithmetic sequence?

A

-3.5, -7.0, -10.5

B

-3.5, 7.0, -14.0

C

3.5, -7.0, 10.5

D

3.5, 7.0, 14.0

Reveal Answer
A

-3.5, -7.0, -10.5

Correct Answer

This is correct because an arithmetic sequence has a constant difference between consecutive terms. Here, the common difference is 3.5-3.5 since 7.0(3.5)=3.5-7.0 - (-3.5) = -3.5 and 10.5(7.0)=3.5-10.5 - (-7.0) = -3.5.

B

-3.5, 7.0, -14.0

This is incorrect because the difference between terms is not constant. This is actually a geometric sequence where each term is multiplied by a common ratio of 2-2.

C

3.5, -7.0, 10.5

This is incorrect because the difference between consecutive terms changes from 10.5-10.5 to 17.517.5. An arithmetic sequence must have a constant common difference.

D

3.5, 7.0, 14.0

This is incorrect because the difference between terms is not constant (7.03.5=3.57.0 - 3.5 = 3.5, but 14.07.0=7.014.0 - 7.0 = 7.0). This represents a geometric sequence with a common ratio of 22.

Q6
2025
QCAA
Paper 1
1 mark
Q6
1 mark

A ball is dropped from a height of 25.6 m. After each bounce, the ball rebounds to 75% of its previous height.

Which option shows the ball's height after the third bounce?

A

8.1 m

B

10.8 m

C

14.4 m

D

19.2 m

Reveal Answer
A

8.1 m

This is incorrect because it represents the height of the ball after the fourth bounce (25.6×0.754=8.125.6 \times 0.75^4 = 8.1 m), not the third.

B

10.8 m

Correct Answer

This is correct. The height after the third bounce is found by multiplying the initial height by the rebound rate cubed: 25.6×0.753=10.825.6 \times 0.75^3 = 10.8 m.

C

14.4 m

This is incorrect because it represents the height of the ball after the second bounce (25.6×0.752=14.425.6 \times 0.75^2 = 14.4 m).

D

19.2 m

This is incorrect because it represents the height of the ball after the first bounce (25.6×0.75=19.225.6 \times 0.75 = 19.2 m).

Q3
2020
QCAA
Paper 1
1 mark
Q3
1 mark

For the sequence 4, 2, 0, –2, –4 … the common difference is

A

4

B

2

C

–2

D

–4

Reveal Answer
A

4

This is the first term of the sequence (a1a_1), not the common difference between terms.

B

2

This value is obtained by subtracting the second term from the first (424 - 2), but the formula for common difference is an+1ana_{n+1} - a_n (second term minus first term).

C

–2

Correct Answer

The common difference is calculated by subtracting a term from the subsequent term: 24=22 - 4 = -2.

D

–4

This is the fifth term of the sequence, not the constant value added to each term to get the next.

Q17
2022
SCSA
Paper 2
8 marks
Q17

Indie was in a line with 24 other people for a slide at a water park. She noticed that the approximate number of people (PP) in the line for the slide increased by 1.5% every minute (mm).

Q17a
2 marks

Write an exponential equation in the form P=armP = ar^m to represent this situation.

Reveal Answer

P=25×1.015mP = 25 \times 1.015^m

Marking Criteria
DescriptorMarks

states correct value of aa

1

states correct value of rr

1
Q17b
2 marks

Determine the approximate number of people in the line after 2 hours.

Reveal Answer

P=25×1.015120P = 25 \times 1.015^{120}
P=149.23P = 149.23

~149 people in line.

Marking Criteria
DescriptorMarks

states correct value

1

recognises integer value required

1
Q17c
4 marks

After 3 hours, the line started to decrease by 1% per minute.

Using this new information, calculate the approximate number of people in line, 5 hours after Indie initially lined up.

Reveal Answer

P=25×1.015180P = 25 \times 1.015^{180}
P=364.61365P = 364.61 \approx365

P=365×0.99mP = 365 \times 0.99^m
P=365×0.99120P = 365 \times 0.99^{120}
P=109.3P = 109.3

~109 people in line.

Marking Criteria
DescriptorMarks

calculates PP for m=180m = 180

1

states new ratio of 0.99

1

identifies m=120m = 120

1

uses equation to calculate P109P \sim 109

1
Q16
2024
QCAA
Paper 1
3 marks
Q16

The number of seats in each row of a theatre forms the terms of the arithmetic sequence
tn+1=tn+8t_{n+1} = t_n + 8, where t1=25t_1 = 25.

Q16a
1 mark

How many seats are in the second row of the theatre?

Reveal Answer

t2=t1+8t_2 = t_1 + 8
=25+8= 25 + 8
=33= 33

The second row of the theatre has 33 seats.

Marking Criteria
DescriptorMarks

correctly determines the number of seats in the second row

1
Q16b
2 marks

Complete the table and then calculate the total number of seats in the first four rows of the theatre.

Row1234
Number of seats    
Reveal Answer

t3=t2+8t4=t3+8t_3 = t_2 + 8 \quad t_4 = t_3 + 8
=33+8=41+8= 33 + 8 \quad = 41 + 8
=41=49= 41 \quad = 49

Row1234
Number of seats25334149

Total number of seats in first four rows of the theatre
=25+33+41+49= 25 + 33 + 41 + 49
=148= 148

Marking Criteria
DescriptorMarks

correctly completes the table to display the first four terms

1

calculates total number of seats in first four rows

1
Q6
2023
QCAA
Paper 1
1 mark
Q6
1 mark

In January 2022, 40 fish were released into a new dam that has the capacity to support 10 000 fish. It is predicted that the dam will reach its capacity in January 2030 if the fish population doubles every year.
Which sequence rule models the prediction?

A

tn=t1r(n1)t_n = t_1 r^{(n-1)}, where t1=40,r=2,n=8t_1 = 40, r = 2, n = 8

B

tn=t1r(n1)t_n = t_1 r^{(n-1)}, where t1=40,r=2,n=9t_1 = 40, r = 2, n = 9

C

tn=t1+(n1)dt_n = t_1 + (n-1)d, where t1=40,d=2,n=8t_1 = 40, d = 2, n = 8

D

tn=t1+(n1)dt_n = t_1 + (n-1)d, where t1=40,d=2,n=9t_1 = 40, d = 2, n = 9

Reveal Answer
A

tn=t1r(n1)t_n = t_1 r^{(n-1)}, where t1=40,r=2,n=8t_1 = 40, r = 2, n = 8

This option correctly identifies the geometric nature of the growth, but the value for nn is incorrect. Since January 2022 is the 1st term (n=1n=1), January 2030 is 8 years later, making it the 9th term (n=9n=9).

B

tn=t1r(n1)t_n = t_1 r^{(n-1)}, where t1=40,r=2,n=9t_1 = 40, r = 2, n = 9

Correct Answer

The population doubles every year, requiring a geometric sequence with r=2r=2 and t1=40t_1=40. Counting inclusively from January 2022 (n=1n=1) to January 2030 results in n=9n=9 terms.

C

tn=t1+(n1)dt_n = t_1 + (n-1)d, where t1=40,d=2,n=8t_1 = 40, d = 2, n = 8

This option uses the arithmetic sequence formula, which models adding a fixed amount (dd) each year. Since the population doubles (multiplies), a geometric formula is required.

D

tn=t1+(n1)dt_n = t_1 + (n-1)d, where t1=40,d=2,n=9t_1 = 40, d = 2, n = 9

This is incorrect because it applies an arithmetic rule (tn=t1+(n1)dt_n = t_1 + (n-1)d). Doubling represents exponential growth, which must be modeled by a geometric sequence.

Q1
2021
SCSA
Paper 1
5 marks
Q1

Hanai is a successful college basketball player. His coach has warned him that he will lose his scholarship if he scores 54% or below on a weekly assessment. On his first three weekly assessments he scored 84%, 81% and 78% respectively.

Assume Hanai's weekly assessments continue to follow this pattern.

Q1a
2 marks

Deduce a rule for the nthn^{th} term of this sequence.

Reveal Answer

Tn=84+(n1)(3)=873nT_n = 84 + (n-1)(-3) = 87 - 3n

Marking Criteria
DescriptorMarks

correctly identifies an arithmetic sequence

1

correctly states the rule for the nthn^{\text{th}} term

1
Q1b
1 mark

Determine Hanai's score on his sixth weekly assessment.

Reveal Answer

T6=873(6)=69T_6 = 87 - 3(6) = 69

Therefore, he gets 69% on his sixth assessment

Marking Criteria
DescriptorMarks

calculates the correct value

1
Q1c
2 marks

Predict when Hanai will lose his scholarship.

Reveal Answer

54=873n3n=33n=11\begin{align*} 54 &= 87 - 3n\\ 3n &= 33\\ n &= 11 \end{align*}

Therefore, Hanai will lose his scholarship after the 11th11^{\text{th}} weekly assessment

Marking Criteria
DescriptorMarks

substitutes 54 correctly

1

identifies correct weekly assessment

1
Q7
2021
QCAA
Paper 2
6 marks
Q7
6 marks

The table shows the total number of times a new song is played on a music service in the days following its first release.

Number of days since first release5101520
Total number of times played ('000s)8121827

The songwriter is paid 0.175 cents every time their song is played and will be paid after 60 days. They predict that by that time, they will be owed at least $1000.

Given that the number of times the song is played is increasing exponentially, evaluate the reasonableness of this prediction.

Reveal Answer

Let n=# of days5n = \frac{\# \text{ of days}}{5}
Let tn=t_n = the total number of plays

t1=8\therefore t_1 = 8

r=128r = \frac{12}{8}
=1.5= 1.5

tn=8×1.5(n1)\therefore t_n = 8 \times 1.5^{(n-1)}
At 60 days
n=605n = \frac{60}{5}
=12= 12

Total number of plays (in 1000s)
t12=8×1.511\therefore t_{12} = 8 \times 1.5^{11}
=691.98= 691.98

Total predicted income
Income =0.175×691980= 0.175 \times 691\,980
=121096.5 cents= 121\,096.5 \text{ cents}
=1210.97= $1210.97

At least 1000$1000 is a reasonable prediction if plays continue as a geometric progression.

Marking Criteria
DescriptorMarks

correctly defines the variables

1

correctly determines the parameter rr

1

correctly determines a geometric (exponential) model

1

determines total number of plays

1

determines income

1

evaluates reasonableness of solution

1
Q20
2021
QCAA
Paper 1
4 marks
Q20
4 marks

A farmer bought a tractor for $45 100 at the start of 2012. It depreciates by $2700 each year.
Identify and use a mathematical model to determine the value of the tractor at the start of 2021.

Reveal Answer

t1=45100t_1 = 45\,100
d=2700d = -2700
n=10n = 10
tn=?t_n = ?

tn=t1+(n1)dt_n = t_1 + (n-1)d
tn=451002700(101)\therefore t_n = 45\,100 - 2700(10-1)
=20800\therefore = 20\,800

The tractor will be worth $20 800.

Marking Criteria
DescriptorMarks

correctly identifies the model

1

correctly identifies the parameters t1t_1, dd and nn

1

substitutes values into appropriate model

1

determines value of tractor, including units

1
Q21
2023
VCAA
Paper 1
1 mark
Q21
1 mark

Use the following information to answer the question.

For taxation purposes, Audrey depreciates the value of her $3000 computer over a four-year period. At the end of the four years, the value of the computer is $600.

If Audrey uses reducing balance depreciation, the depreciation rate, per annum is closest to

A

10%

B

15%

C

20%

D

25%

E

33%

Reveal Answer
A

10%

Incorrect. A 10% reducing balance rate would leave a final value of 3000(10.10)4=1968.303000(1 - 0.10)^4 = $1968.30, which is much higher than the actual final value of $600.

B

15%

Incorrect. Applying a 15% reducing balance rate gives a final value of 3000(10.15)4=1566.023000(1 - 0.15)^4 = $1566.02, which does not match the target of $600.

C

20%

Incorrect. This represents the straight-line depreciation rate, where the computer loses $600 per year (20% of $3000), rather than the reducing balance rate.

D

25%

Incorrect. A 25% reducing balance rate results in a final value of 3000(10.25)4=949.223000(1 - 0.25)^4 = $949.22, which is still higher than $600.

E

33%

Correct Answer

Correct. Using the reducing balance formula Vn=V0(1r)nV_n = V_0(1 - r)^n, we set up the equation 600=3000(1r)4600 = 3000(1 - r)^4. Solving for rr yields r=10.240.331r = 1 - \sqrt[4]{0.2} \approx 0.331, which is closest to 33%.

Q17
2025
VCAA
Paper 1
1 mark
Q17
1 mark

Dani invests $4000 for three years.

The account earns simple interest at 4% per annum.

The balance in the account after three years can be calculated using

A

4000×1.0434000 \times 1.04^3

B

3(1.04×4000)3(1.04 \times 4000)

C

4000+(0.04×4000)34000 + (0.04 \times 4000)^3

D

4000+3(0.04×4000)4000 + 3(0.04 \times 4000)

Reveal Answer
A

4000×1.0434000 \times 1.04^3

Incorrect. This formula calculates compound interest (A=P(1+r)tA = P(1 + r)^t), but the question specifies simple interest.

B

3(1.04×4000)3(1.04 \times 4000)

Incorrect. This calculates three times the total balance after one year, which incorrectly triples the initial principal instead of just multiplying the interest by 3.

C

4000+(0.04×4000)34000 + (0.04 \times 4000)^3

Incorrect. This cubes the interest earned in a single year instead of multiplying it by the number of years (t=3t=3).

D

4000+3(0.04×4000)4000 + 3(0.04 \times 4000)

Correct Answer

Correct. The total balance is the principal plus the simple interest (A=P+PrtA = P + Prt), which is calculated as $4000 plus 3 years of 4% interest on the principal.

Q9
2024
QCAA
Paper 1
1 mark
Q9
1 mark

Determine the 4th term for the geometric sequence that begins 1000, -900, ...

A

729

B

700

C

-700

D

-729

Reveal Answer
A

729

This incorrect answer ignores the negative sign of the common ratio. Since r=0.9r = -0.9 and the exponent is odd (n1=3n-1=3), the 4th term must be negative.

B

700

This option incorrectly treats the sequence as arithmetic (subtracting 100 each time) rather than geometric, and ignores the negative sign of the second term.

C

-700

This answer suggests an arithmetic pattern rather than a geometric one. In a geometric sequence, terms are multiplied by a common ratio, not added or subtracted by a constant.

D

-729

Correct Answer

First, determine the common ratio r=9001000=0.9r = \frac{-900}{1000} = -0.9. Then, calculate the 4th term using a4=1000(0.9)3=729a_4 = 1000 \cdot (-0.9)^3 = -729.

Q18
2020
QCAA
Paper 1
4 marks
Q18

Exhibition organisers believe that the number of attendees increases each day as an arithmetic sequence. The organisers know that 353 people attended the first day and 439 people attended the third day.

Q18a
2 marks

Determine the common difference.

Reveal Answer

Arithmetic sequence
t1=353t_1 = 353
t3=439t_3 = 439

Find dd
t3=t1+2dt_3 = t_1 + 2d
439=353+2d439 = 353 + 2d
86=2d86 = 2d
43=d43 = d

Marking Criteria
DescriptorMarks

Correctly provides mathematical reasoning to support the answer

1

Correctly determines the common difference

1
Q18b
2 marks

Use the result from 18a) to predict the number of people who will attend the sixth day.

Reveal Answer

Find t6t_6
t6=t1+5dt_6 = t_1 + 5d
=353+5×43= 353 + 5 \times 43
=568= 568

They would expect 568 people to attend the sixth day.

Marking Criteria
DescriptorMarks

Substitutes into an appropriate rule

1

Determines value

1

Frequently Asked Questions

How many QCAA General Mathematics questions cover Growth and decay in sequences?
AusGrader has 112 QCAA General Mathematics questions on Growth and decay in sequences, all with instant AI grading and detailed marking feedback.

Ready to practise QCAA General Mathematics?

Get instant AI feedback on past exam questions, aligned to the syllabus

Start Practising Free