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WACE Mathematics Applications — Growth and decay in sequences Questions & 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

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

Marking Criteria

Descriptor	Marks
states correct recursive rule	1
states correct first term	1

Q3b

2 marks

Deduce a simplified rule for the $n$ th term of the monthly wage.

Reveal Answer

$T_n = 4000 + (n-1)(50)$
$T_n = 3950 + 50n$

Marking Criteria

Descriptor	Marks
uses correct arithmetic formula	1
gives correct simplified rule for the $n^{\text{th}}$ term	1

Q3c

2 marks

Determine the monthly wage for December 2020.

Reveal Answer

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

Therefore, the monthly wage for December 2020 is ＄4550

Marking Criteria

Descriptor	Marks
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

$12 \times 4 = 48$
$T_{48} = 3950 + 50(48) = 6350$

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

Marking Criteria

Descriptor	Marks
correctly calculates $T_{48}$	1
calculates correct term for March 2024	1
states correct solution for wage	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
$t_{n+1} = t_n + 8$ , where $t_1 = 25$ .

Q16a

1 mark

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

Reveal Answer

$t_2 = t_1 + 8$
$= 25 + 8$
$= 33$

The second row of the theatre has 33 seats.

Marking Criteria

Descriptor	Marks
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.

Row	1	2	3	4
Number of seats

Reveal Answer

$t_3 = t_2 + 8 \quad t_4 = t_3 + 8$
$= 33 + 8 \quad = 41 + 8$
$= 41 \quad = 49$

Row	1	2	3	4
Number of seats	25	33	41	49

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

Marking Criteria

Descriptor	Marks
correctly completes the table to display the first four terms	1
calculates total number of seats in first four rows	1

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 ( $a_1$ ), not the common difference between terms.

B

2

This value is obtained by subtracting the second term from the first ( $4 - 2$ ), but the formula for common difference is $a_{n+1} - a_n$ (second term minus first term).

C

–2

Correct Answer

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

D

–4

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

Q6

2025

SCSA

Paper 2

13 marks

Q6

Josh lives in Sydney and Simon lives in Perth. They each decide to drive to Alice Springs to meet before driving together to Darwin. Simon and Josh will both leave for Alice Springs on Saturday, 2 January.

Due to traffic in Sydney, Josh can only drive 250 km on the first day. After this he plans to drive 3 km more each day than the previous day.

Let $n$ represent the number of days spent driving.

Q6c

The distance Simon drives each day can be represented by the rule $T_n = 220(1.06)^{n-1}$ .

Q6e

The approximate distance from Sydney to Alice Springs is 2770 km and from Perth to Alice Springs is 2550 km.

Q6a

2 marks

Write a recursive rule to model the distance Josh drives each day.

Reveal Answer

$T_{n+1} = T_n + 3, \quad T_1 = 250$

Marking Criteria

Descriptor	Marks
states correct rule	1
states correct value for $T_1$	1

Q6b

1 mark

Calculate how far Josh is away from Sydney by the end of day four.

Reveal Answer

$250 + 253 + 256 + 259 = 1018$ kilometres

Marking Criteria

Descriptor	Marks
calculates correct distance	1

Q6c

2 marks

Interpret what the numbers 220 and 1.06 represent in the context of the question.

Reveal Answer

The 220 represents the distance travelled in kilometres by Simon by the end of day one.
1.06 represents a 6% increase in distance travelled each day.

Marking Criteria

Descriptor	Marks
correctly interprets what the 220 represents in the context of the question	1
correctly interprets what the 1.06 represents in the context of the question	1

Q6d

3 marks

Determine the day and date when Simon drove 330.8 km.

Reveal Answer

\begin{align*} 330.8 &= 220(1.06)^{n-1}\\ n &= 8 \end{align*}

Therefore Simon travels 330.8 km on day 8 which is Saturday, 9 January.

Marking Criteria

Descriptor	Marks
solves correctly for $n$ or states $T_8 = 330.8$	1
states correct day	1
states correct date	1

Q6e

3 marks

Determine who is the first to arrive in Alice Springs. Justify your answer.

Reveal Answer

Josh $\rightarrow$ total distance after 10 days = 2635 kilometres
total distance after 11 days = 2915 kilometres
Therefore Josh will arrive on the $11^{\text{th}}$ day.

Simon $\rightarrow$ total distance after 9 days = 2528.09 kilometres
total distance after 10 days = 2899.77 kilometres
Therefore Simon will arrive on the $10^{\text{th}}$ day.

Simon will arrive in Alice springs first.

Marking Criteria

Descriptor	Marks
determines Simon's correct day of arrival	1
determines Josh's correct day of arrival	1
concludes who will arrive first with justification based on the day of arrival	1

Q6f

2 marks

After Josh and Simon meet in Alice Springs, they drive together to Darwin. The distance they travel each day can be represented by the rule $T_{n+1} = aT_n + b$ .

Given they travel 380 km on day one, 309 km on day two and 269.95 km on day three, determine the value of $a$ and $b$ .

Reveal Answer

$309 = a(380) + b$ ...(1)
$269.95 = a(309) + b$ ...(2)

$a = 0.55, \; b = 100$

Marking Criteria

Descriptor	Marks
determines the correct value for $a$	1
determines the correct value for $b$	1

Q17

2024

VCAA

Paper 1

1 mark

Q17

1 mark

A first-order linear recurrence relation of the form

$u_0 = a, \quad u_{n+1} = R u_n + d$

generates the terms of a sequence. A geometric sequence will be generated if

A

$R = 1$ and $d = -1$

B

$R = 1$ and $d = 1$

C

$R = 4$ and $d = -1$

D

$R = 2$ and $d = 0$

Reveal Answer

A

$R = 1$ and $d = -1$

If $R = 1$ and $d = -1$ , the relation becomes $u_{n+1} = u_n - 1$ . This generates an arithmetic sequence with a common difference of $-1$ , not a geometric sequence.

B

$R = 1$ and $d = 1$

If $R = 1$ and $d = 1$ , the relation becomes $u_{n+1} = u_n + 1$ . This generates an arithmetic sequence with a common difference of $1$ .

C

$R = 4$ and $d = -1$

If $R = 4$ and $d = -1$ , the relation becomes $u_{n+1} = 4u_n - 1$ . Because $d \neq 0$ , this generates a sequence that is neither purely arithmetic nor purely geometric.

D

$R = 2$ and $d = 0$

Correct Answer

A geometric sequence requires a constant ratio between consecutive terms, meaning it must be in the form $u_{n+1} = r u_n$ . Setting $d = 0$ and $R = 2$ gives $u_{n+1} = 2u_n$ , which perfectly fits the definition of a geometric sequence.

Q8

2024

QCAA

Paper 1

1 mark

Q8

1 mark

After $n$ bounces, the rebound height (cm) of a ball, $t_n$ , is modelled by the rule $t_n = 240 \times 0.5^{(n-1)}$ .
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 ( $t_2 = 120$ ) or the difference between the first and second bounces, rather than the difference between the first and third.

C

180

Correct Answer

First calculate $t_1 = 240 \times 0.5^0 = 240$ and $t_3 = 240 \times 0.5^2 = 60$ . The difference is $240 - 60 = 180$ .

D

210

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

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 ( $P$ ) in the line for the slide increased by 1.5% every minute ( $m$ ).

Q17a

2 marks

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

Reveal Answer

$P = 25 \times 1.015^m$

Marking Criteria

Descriptor	Marks
states correct value of $a$	1
states correct value of $r$	1

Q17b

2 marks

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

Reveal Answer

$P = 25 \times 1.015^{120}$
$P = 149.23$

~149 people in line.

Marking Criteria

Descriptor	Marks
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

$P = 25 \times 1.015^{180}$
$P = 364.61 \approx365$

$P = 365 \times 0.99^m$
$P = 365 \times 0.99^{120}$
$P = 109.3$

~109 people in line.

Marking Criteria

Descriptor	Marks
calculates $P$ for $m = 180$	1
states new ratio of 0.99	1
identifies $m = 120$	1
uses equation to calculate $P \sim 109$	1

Q17

2023

VCAA

Paper 1

1 mark

Q17

1 mark

A sequence of numbers is generated by the recurrence relation shown below.

$T_0 = 5, \quad T_{n+1} = -T_n$

The value of $T_2$ is

A

$-10$

B

$-5$

C

0

D

5

E

10

Reveal Answer

A

$-10$

This is incorrect. This value might be obtained by subtracting 5 at each step, but the recurrence relation multiplies the previous term by $-1$ .

B

$-5$

This is incorrect. $-5$ is the value of $T_1$ , not $T_2$ . You must apply the recurrence relation one more time to find $T_2$ .

C

0

This is incorrect. This might result from adding $T_0$ and $T_1$ , but the relation requires simply negating the previous term.

D

5

Correct Answer

This is correct. Using the recurrence relation, $T_1 = -T_0 = -5$ . Applying it again gives $T_2 = -T_1 = -(-5) = 5$ .

E

10

This is incorrect. This assumes the sequence increases by 5 at each step, rather than alternating signs.

Q8

2020

SCSA

Paper 2

9 marks

Q8

A farmer has a large lake on his farm and has started stocking it with fish of a variety that will flourish in the conditions in this lake. Monitoring has shown that the number of adult fish is increasing at a consistent rate of 9% per month and at the beginning of 2020 the lake holds 660 of the adult fish.

Q8c

The farmer plans to allow the general public to pay to fish in the lake. This will commence at the beginning of the next month after the adult fish population first reaches 4000.

Q8a

2 marks

Write a recursive rule to give the number of adult fish in the lake at the end of each month from the beginning of 2020.

Reveal Answer

$T_{n+1} = 1.09T_n \qquad T_0 = 660$

Marking Criteria

Descriptor	Marks
correctly states recursive rule	1
correctly states $T_0$	1

Q8b

2 marks

Deduce a rule for the $n^{\text{th}}$ term of this sequence.

Reveal Answer

$A_n = 660 \times 1.09^n$

Marking Criteria

Descriptor	Marks
gives formula in exponential form	1
states correct rule	1

Q8c

2 marks

Determine how many months after the beginning of 2020 fishing will commence.

Reveal Answer

$4000 = 660 \times 1.09^n \implies n = 20.9$
Therefore fishing will commence 21 months after the beginning of 2020

Marking Criteria

Descriptor	Marks
correctly solves for $n$	1
correctly states correct number of months after the beginning of 2020	1

Q8d

3 marks

The farmer wishes to maintain a steady state in the adult fish population once fishing commences. Calculate how many adult fish can be taken from the lake each month.

Reveal Answer

$T_{21} = 4031$
$T_{n+1} = 1.09T_n - x$

$4031 = 1.09 \times 4031 - x \implies x = 362.79$

362 fish per month

Marking Criteria

Descriptor	Marks
correctly determines $T_{21} = 4031$	1
correctly determines monthly increase	1
correctly rounds down for number of fish	1

Q1

2025

SCSA

Paper 1

10 marks

Q1

When people become ill, their body responds with a change to their white blood cell count (WBCC) measured in '000s cells per microlitre (cells/μL).

The table below shows a person's white blood cell count after contracting an illness.
Let $n$ represent the number of hours after contracting an illness.

Time ( $n$ hours)	1	2	3	4	5	6
WBCC '000s (cells/μL)	8.4	8.7	9

Q1e

After an illness is diagnosed, a person receives the appropriate medication to aid their recovery. An indirect effect of the medication will be an overall reduction in the person's white blood cell count.

A person becomes ill and is given medication. Their white blood cell count ('000s cells/μL) follows the recursive rule $T_m = -0.5T_{m-1} + 20, \quad T_0 = 20$ , where $m$ is the number of hours after the person is given the medication.

Q1a

1 mark

The person's white blood cell count is increasing at a constant rate each hour. Show that this rate is 300 cells/μL.

Reveal Answer

$8700 - 8400 = 300$

Therefore the constant rate is 300 cells/ $\mu$ L.

Marking Criteria

Descriptor	Marks
correctly shows that the constant rate is 300 cells/ $\mu$ L	1

Q1b

1 mark

Complete the table above.

Reveal Answer

Time ( $n$ hours)	1	2	3	4	5	6
WBCC '000s (cells/ $\mu$ L)	8.4	8.7	9	9.3	9.6	9.9

Marking Criteria

Descriptor	Marks
determines all correct entries	1

Q1c

2 marks

Determine a simplified rule for the $n^{\text{th}}$ term to model the person's white blood cell count after contracting an illness.

Reveal Answer

$T_n = 8.4 + (n - 1)(0.3)$
$T_n = 8.1 + 0.3n$

Marking Criteria

Descriptor	Marks
uses the rule for an arithmetic sequence and substitutes correct $a$ and $d$ values	1
correctly simplifies the rule	1

Q1d

2 marks

Determine the white blood cell count in the person's body after 12 hours, if the white blood cell count maintains the same rate of increase.

Reveal Answer

$T_{12} = 8.1 + 0.3(12)$
$T_{12} = 11.7$

Therefore 11 700 cells/ $\mu$ L.

Marking Criteria

Descriptor	Marks
correctly determines the twelfth term	1
states correct white blood cell count in cells/ $\mu$ L	1

Q1e (i)

2 marks

Determine the white blood cell count in this person at the end of the second hour.

Reveal Answer

$T_1 = (-0.5)(20) + 20 = 10$
$T_2 = (-0.5)(10) + 20 = 15$

Therefore the white blood cell count in the body is 15 000 cells/ $\mu$ L at the end of the second hour.

Marking Criteria

Descriptor	Marks
determines correct value for $T_1$	1
states correct white blood cell count in cells/ $\mu$ L	1

Q1e (ii)

2 marks

Determine if the given medication, in the long run, can ensure this person's white blood cell count does not fall below 13 000 cells per microlitre.

Reveal Answer

$x = -0.5x + 20 \Rightarrow x = \frac{20}{1.5} = \frac{200}{15} = \frac{40}{3} = 13\frac{1}{3} \approx 13.3$

i.e. approximately 13 300 cells/ $\mu$ L, which is greater than 13 000

Therefore the given medication will ensure the person's white blood cell count does not fall below 13 000 cells per microlitre.

Marking Criteria

Descriptor	Marks
determines correct steady-state value	1
correctly justifies medication will ensure the person's white blood cell count does not fall below 13 000 cells per microlitre	1

Q5

2024

VCAA

Paper 2

4 marks

Q5

Emi operates a mobile dog-grooming business.
The value of her grooming equipment will depreciate.
Based on average usage, a rule for the value, in dollars, of the equipment, $V_n$ , after $n$ weeks is

$V_n = 15000 - 60n$

Assume that there are exactly 52 weeks in a year.

Q5a

1 mark

By what amount, in dollars, does the value of the grooming equipment depreciate each week?

Reveal Answer

＄60

Marking Criteria

Descriptor	Marks
States the correct depreciation amount per week, ＄60.	1

Q5b

1 mark

Emi plans to replace the grooming equipment after four years.

What will be its value, in dollars, at this time?

Reveal Answer

＄2520

Marking Criteria

Descriptor	Marks
Calculates the correct value after four years, ＄2520.	1

Q5c

1 mark

$V_n$ is the value of the grooming equipment, in dollars, after $n$ weeks.

Write a recurrence relation in terms of $V_0$ , $V_{n+1}$ and $V_n$ that can model this value from one week to the next.

Reveal Answer

$V_0 = 15\,000, \quad V_{n+1} = V_n - 60$

Marking Criteria

Descriptor	Marks
Writes the correct recurrence relation, $V_0 = 15\,000, V_{n+1} = V_n - 60$ .	1

Q5d

1 mark

The value of the grooming equipment decreases from one year to the next by the same percentage of the original ＄15000 value.

What is this annual flat rate percentage?

Reveal Answer

20.8%

Marking Criteria

Descriptor	Marks
Calculates the correct annual flat rate percentage, 20.8%.	1

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 $n^{th}$ term of this sequence.

Reveal Answer

$T_n = 84 + (n-1)(-3) = 87 - 3n$

Marking Criteria

Descriptor	Marks
correctly identifies an arithmetic sequence	1
correctly states the rule for the $n^{\text{th}}$ term	1

Q1b

1 mark

Determine Hanai's score on his sixth weekly assessment.

Reveal Answer

$T_6 = 87 - 3(6) = 69$

Therefore, he gets 69% on his sixth assessment

Marking Criteria

Descriptor	Marks
calculates the correct value	1

Q1c

2 marks

Predict when Hanai will lose his scholarship.

Reveal Answer

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

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

Marking Criteria

Descriptor	Marks
substitutes 54 correctly	1
identifies correct weekly assessment	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

$t_1 = 45\,100$
$d = -2700$
$n = 10$
$t_n = ?$

$t_n = t_1 + (n-1)d$
$\therefore t_n = 45\,100 - 2700(10-1)$
$\therefore = 20\,800$

The tractor will be worth ＄20 800.

Marking Criteria

Descriptor	Marks
correctly identifies the model	1
correctly identifies the parameters $t_1$ , $d$ and $n$	1
substitutes values into appropriate model	1
determines value of tractor, including units	1

Q18

2025

VCAA

Paper 1

1 mark

Q18

1 mark

A recurrence relation is of the form

$u_0 = a, \quad u_{n+1} = R u_n + d$

If $a > 0$ , $R = 0.5$ and $d = 0$ , the sequence generated will be

A

arithmetic and increasing.

B

arithmetic and decreasing.

C

geometric and increasing.

D

geometric and decreasing.

Reveal Answer

A

arithmetic and increasing.

The sequence is not arithmetic because each term is multiplied by a constant $R$ , rather than having a constant added to it. It is also decreasing, not increasing.

B

arithmetic and decreasing.

While the sequence is decreasing, it is not arithmetic. An arithmetic sequence requires a constant difference between terms, which would mean $R = 1$ and $d \neq 0$ .

C

geometric and increasing.

The sequence is geometric, but because the common ratio $R = 0.5$ is less than 1 and the initial term $a$ is positive, the sequence is decreasing, not increasing.

D

geometric and decreasing.

Correct Answer

Substituting the given values yields $u_{n+1} = 0.5 u_n$ , which defines a geometric sequence. Since the initial term $a > 0$ and the common ratio $0.5$ is between 0 and 1, the terms will get progressively smaller, making it decreasing.

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

$t_n = t_1 r^{(n-1)}$ , where $t_1 = 40, r = 2, n = 8$

B

$t_n = t_1 r^{(n-1)}$ , where $t_1 = 40, r = 2, n = 9$

C

$t_n = t_1 + (n-1)d$ , where $t_1 = 40, d = 2, n = 8$

D

$t_n = t_1 + (n-1)d$ , where $t_1 = 40, d = 2, n = 9$

Reveal Answer

A

$t_n = t_1 r^{(n-1)}$ , where $t_1 = 40, r = 2, n = 8$

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

B

$t_n = t_1 r^{(n-1)}$ , where $t_1 = 40, r = 2, n = 9$

Correct Answer

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

C

$t_n = t_1 + (n-1)d$ , where $t_1 = 40, d = 2, n = 8$

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

D

$t_n = t_1 + (n-1)d$ , where $t_1 = 40, d = 2, n = 9$

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

SCSA Mathematics Applications Growth and decay in sequences

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