Study Mode: This practice Application Task covers arithmetic sequences and series, geometric sequences and series, recurrence relations, and applications including financial modelling. Work through each part with your CAS calculator and textbook nearby. Collapse the solutions and try each question before checking.

Specialist Mathematics Unit 1

SAC 1: Application Task

Unit:Specialist Mathematics Unit 1 Task format:Application Task Working time:50 minutes Reading time:5 minutes Total marks:50 marks CAS calculator:Permitted

Note: This is a practice paper. Your school's SAC may differ in length, marks, and time allocation. Some schools include a tech-free component. The format and difficulty are calibrated to be representative.

Scenario: Sunset Beach Coastal Conservation Project

A conservation team monitors coastal erosion at Sunset Beach and plans a revegetation project. They use sequences and series to model beach width changes, plant growth, and project finances.

Part A: Beach Monitoring (Arithmetic Sequences & Series)

12 marks

Beach width measurements and sand erosion are monitored using arithmetic sequences to model consistent monthly changes.

Question 14 marks

Beach width measurements (metres) taken at the start of each month: 48, 45, 42, 39, ...

a. Show this is an arithmetic sequence and state the common difference. [1 mark]

b. Write a formula for the width $w_n$ at the start of month $n$. [1 mark]

c. Find the width at the start of month 12. [1 mark]

d. Find the month when the width first falls below 15 metres. [1 mark]

Have you genuinely attempted this question on paper first?

a. [1 mark]

Working: $d = 45 - 48 = -3$. Check: $42 - 45 = -3$, $39 - 42 = -3$. Common difference is constant, so the sequence is arithmetic.

$d = -3$ (decreasing by 3 metres per month)

1 mark for showing constant difference and stating $d = -3$.

b. [1 mark]

$w_n = 48 + (n-1)(-3) = 51 - 3n$

c. [1 mark]

$w_{12} = 51 - 3(12) = 51 - 36 = 15$

15 metres

d. [1 mark]

$51 - 3n < 15 \Rightarrow 3n > 36 \Rightarrow n > 12$. So month 13.

Month 13 ($w_{13} = 51 - 39 = 12$ metres)

How did you go? (out of 4)

Question 24 marks

Sand eroded each month (cubic metres): 100 in month 1, increasing by 15 cubic metres each month.

a. Find the amount of sand eroded in month 10. [1 mark]

b. Find the total sand eroded over the first 12 months. [2 marks]

c. In which month does the cumulative erosion first exceed 5000 cubic metres? [1 mark]

a. [1 mark]

$t_{10} = 100 + 9(15) = 100 + 135 = 235$

235 cubic metres

b. [2 marks]

$S_{12} = \frac{12}{2} \times (2(100) + 11(15)) = 6 \times (200 + 165) = 6 \times 365 = 2190$

2190 cubic metres

1 mark for correct substitution into sum formula. 1 mark for correct answer.

c. [1 mark]

$S_n > 5000$. Using $S_n = \frac{n}{2}(200 + (n-1)(15)) = \frac{n}{2}(185 + 15n)$. Solve $\frac{n}{2}(185 + 15n) = 5000 \Rightarrow 15n^2 + 185n - 10000 = 0 \Rightarrow 3n^2 + 37n - 2000 = 0$. By CAS: $n \approx 20.4$. So month 21. Check: $S_{21} = \frac{21}{2}(185 + 315) = \frac{21}{2} \times 500 = 5250 > 5000$ ✓. $S_{20} = \frac{20}{2}(185 + 300) = 10 \times 485 = 4850 < 5000$ ✓

Month 21

How did you go? (out of 4)

Question 34 marks

A second section of beach has width modelled by the recurrence relation $w_{n+1} = 0.95w_n + 1$, with $w_1 = 40$ metres.

a. Find $w_2$, $w_3$, and $w_4$. Give answers correct to 2 decimal places. [2 marks]

b. Find the steady-state width (where $w_{n+1} = w_n$). [1 mark]

c. Interpret the steady state in the context of the problem. [1 mark]

a. [2 marks]

$w_2 = 0.95(40) + 1 = 38 + 1 = 39$. $w_3 = 0.95(39) + 1 = 37.05 + 1 = 38.05$. $w_4 = 0.95(38.05) + 1 = 36.1475 + 1 = 37.15$

$w_2 = 39$, $w_3 = 38.05$, $w_4 = 37.15$

1 mark for $w_2$ correct. 1 mark for $w_3$ and $w_4$ correct.

b. [1 mark]

Set $w = 0.95w + 1 \Rightarrow 0.05w = 1 \Rightarrow w = 20$

Steady-state width is 20 metres.

c. [1 mark]

The beach width decreases over time but approaches 20 metres. This means natural sand replenishment (the +1 term) eventually balances the erosion (the 0.95 multiplier), and the beach stabilises at 20 metres wide.

How did you go? (out of 4)

Part B: Revegetation (Geometric Sequences & Series)

13 marks

Native plants are introduced with exponential growth patterns. Seedling germination follows a decreasing geometric sequence over seasons.

Question 44 marks

Native spinifex plants are introduced: 80 plants in season 1, with the number increasing by 25% each season.

a. Write an expression for the number of plants $P_n$ in season $n$. [1 mark]

b. Find the number of plants in season 6, correct to the nearest whole number. [1 mark]

c. In which season does the plant count first exceed 1000? [2 marks]

a. [1 mark]

$P_n = 80 \times 1.25^{n-1}$

b. [1 mark]

$P_6 = 80 \times 1.25^5 = 80 \times 3.05176... = 244.14...$

244 plants

c. [2 marks]

$80 \times 1.25^{n-1} > 1000 \Rightarrow 1.25^{n-1} > 12.5$. Taking logarithms: $(n-1) \log(1.25) > \log(12.5) \Rightarrow n - 1 > \frac{\log(12.5)}{\log(1.25)} = 11.33... \Rightarrow n > 12.33$, so $n = 13$. Check: $P_{13} = 80 \times 1.25^{12} \approx 1164 > 1000$ ✓. $P_{12} = 80 \times 1.25^{11} \approx 931 < 1000$ ✓

Season 13

1 mark for setting up inequality. 1 mark for correct answer with verification.

How did you go? (out of 4)

Question 55 marks

New seedlings germinate each season: 200 in season 1, 150 in season 2, 112.5 in season 3, ...

a. Show this is a geometric sequence and find the common ratio. [1 mark]

b. Find the total number of seedlings germinated over the first 8 seasons. [2 marks]

c. Find the total number of seedlings that will ever germinate (as $n$ approaches infinity). Explain why this sum converges. [2 marks]

a. [1 mark]

$r = \frac{150}{200} = 0.75$. Check: $\frac{112.5}{150} = 0.75$. Constant ratio, so geometric.

Geometric sequence with $r = \frac{3}{4} = 0.75$

b. [2 marks]

$S_8 = \frac{200(1 - 0.75^8)}{1 - 0.75} = \frac{200(1 - 0.1001...)}{0.25} = \frac{200 \times 0.8999...}{0.25} = 200 \times 3.5996... = 719.9$

Approximately 720 seedlings

1 mark for correct formula substitution. 1 mark for correct answer.

c. [2 marks]

$S_\infty = \frac{a}{1-r} = \frac{200}{1 - 0.75} = \frac{200}{0.25} = 800$. The sum converges because $|r| = 0.75 < 1$, so each term is smaller than the last. The terms approach zero, meaning the total approaches a finite limit.

800 seedlings. Converges because $|r| < 1$.

1 mark for correct limiting sum. 1 mark for explaining convergence condition.

How did you go? (out of 5)

Question 64 marks

A geometric sequence has third term 54 and sixth term 2.

a. Find the common ratio. [2 marks]

b. Find the first term. [1 mark]

c. Find the sum of the first 10 terms, correct to 2 decimal places. [1 mark]

a. [2 marks]

$t_3 = ar^2 = 54$ and $t_6 = ar^5 = 2$. Dividing: $\frac{ar^5}{ar^2} = r^3 = \frac{2}{54} = \frac{1}{27}$. $r = \sqrt[3]{\frac{1}{27}} = \frac{1}{3}$

$r = \frac{1}{3}$

1 mark for forming ratio equation. 1 mark for solving for $r$.

b. [1 mark]

$ar^2 = 54 \Rightarrow a\left(\frac{1}{3}\right)^2 = 54 \Rightarrow \frac{a}{9} = 54 \Rightarrow a = 486$

$a = 486$

c. [1 mark]

$S_{10} = \frac{486(1 - (1/3)^{10})}{1 - 1/3} = \frac{486(1 - 1/59049)}{2/3} = 729 \times \frac{59048}{59049} \approx 728.99$

$S_{10} \approx 728.99$

How did you go? (out of 4)

Part C: Project Finances

12 marks

Council funding and equipment depreciation are modelled using arithmetic and geometric sequences to plan long-term project costs and budget allocation.

Question 74 marks

Council allocates $30,000 in Year 1, increasing by $4,000 each year.

a. Write a formula for the funding $F_n$ in year $n$. [1 mark]

b. Find the total funding over 8 years. [2 marks]

c. In which year does total funding first exceed $400,000? [1 mark]

a. [1 mark]

$F_n = 30000 + (n-1)(4000) = 26000 + 4000n$

b. [2 marks]

$S_8 = \frac{8}{2} \times (2(30000) + 7(4000)) = 4 \times (60000 + 28000) = 4 \times 88000 = 352000$

$352,000

1 mark for correct substitution. 1 mark for correct answer.

c. [1 mark]

$S_n > 400000$. Solve $\frac{n}{2}(60000 + (n-1)(4000)) > 400000 \Rightarrow \frac{n}{2}(56000 + 4000n) > 400000 \Rightarrow n(56000 + 4000n) > 800000$. Using CAS: $n \approx 8.78$, so Year 9. $S_9 = \frac{9}{2}(56000 + 36000) = \frac{9}{2} \times 92000 = 414000 > 400000$ ✓

Year 9

How did you go? (out of 4)

Question 84 marks

Equipment purchased for $48,000 depreciates by 15% per year.

a. Write a formula for the value $V_n$ after $n$ years. [1 mark]

b. Find the value after 4 years, correct to the nearest dollar. [1 mark]

c. Find the year when the value first drops below $10,000. [2 marks]

a. [1 mark]

$V_n = 48000 \times 0.85^n$

b. [1 mark]

$V_4 = 48000 \times 0.85^4 = 48000 \times 0.52200625 = 25056.30$

$25,056

c. [2 marks]

$48000 \times 0.85^n < 10000 \Rightarrow 0.85^n < \frac{10000}{48000} = \frac{5}{24}$. Taking logarithms: $n \log(0.85) < \log(5/24) \Rightarrow n > \frac{\log(5/24)}{\log(0.85)} \approx 9.65$. So after 10 years. $V_{10} = 48000 \times 0.85^{10} \approx 9452 < 10000$ ✓. $V_9 = 48000 \times 0.85^9 \approx 11120 > 10000$ ✓

Year 10

1 mark for setting up inequality correctly. 1 mark for correct answer with verification.

How did you go? (out of 4)

Question 94 marks

An invasive weed has population modelled by $W_{n+1} = 1.2W_n - 50$, where $W_0 = 300$.

a. Find $W_1$, $W_2$, and $W_3$. [1 mark]

b. Find the steady-state population. [2 marks]

c. The team increases removal efforts so the model becomes $W_{n+1} = 1.2W_n - k$. Find the value of $k$ that gives a steady state of 200 weeds. [1 mark]

a. [1 mark]

$W_1 = 1.2(300) - 50 = 360 - 50 = 310$. $W_2 = 1.2(310) - 50 = 372 - 50 = 322$. $W_3 = 1.2(322) - 50 = 386.4 - 50 = 336.4$

$W_1 = 310$, $W_2 = 322$, $W_3 = 336.4$

b. [2 marks]

Set $W = 1.2W - 50 \Rightarrow W - 1.2W = -50 \Rightarrow -0.2W = -50 \Rightarrow W = 250$

Steady-state population is 250 weeds.

1 mark for correct equation setup. 1 mark for solving correctly.

c. [1 mark]

$200 = 1.2(200) - k \Rightarrow 200 = 240 - k \Rightarrow k = 40$

$k = 40$

How did you go? (out of 4)

Part D: Connecting Ideas

13 marks

Comparing growth models and proving general formulas deepens understanding of sequence theory and its real-world applications.

Question 104 marks

Two volunteer recruitment models:

Model A: 15 volunteers in week 1, increasing by 5 each week (arithmetic).
Model B: 15 volunteers in week 1, increasing by 30% each week (geometric).

a. Find the number of volunteers in week 10 for each model. [2 marks]

b. Find the total volunteers recruited over 10 weeks for each model. Give answers to the nearest whole number. [2 marks]

a. [2 marks]

Model A: $t_{10} = 15 + 9(5) = 15 + 45 = 60$.
Model B: $t_{10} = 15 \times 1.3^9 = 15 \times 10.6045... \approx 159$

Model A: 60 volunteers. Model B: approximately 159 volunteers.

1 mark for each model correct.

b. [2 marks]

Model A: $S_{10} = \frac{10}{2}(2(15) + 9(5)) = 5(30 + 45) = 5 \times 75 = 375$.
Model B: $S_{10} = \frac{15(1.3^{10} - 1)}{1.3 - 1} = \frac{15(13.786... - 1)}{0.3} = 15 \times 42.62... \approx 639$

Model A: 375 volunteers. Model B: approximately 639 volunteers.

1 mark for each model correct.

How did you go? (out of 4)

Question 114 marks

Prove that for an arithmetic sequence with first term $a$ and common difference $d$, the sum of the first $n$ terms is:

\[S_n = \frac{n}{2}(2a + (n-1)d)\]

[4 marks]

Step 1: Write the sum forwards:
\[S_n = a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d)\]

Step 2: Write the sum in reverse:
\[S_n = (a + (n-1)d) + (a + (n-2)d) + \ldots + (a + d) + a\]

Step 3: Add the two expressions term by term:
\[2S_n = (2a + (n-1)d) + (2a + (n-1)d) + \ldots + (2a + (n-1)d)\]
Each of the $n$ pairs sums to $(2a + (n-1)d)$.
So $2S_n = n(2a + (n-1)d)$

Step 4: Divide both sides by 2:
\[S_n = \frac{n}{2}(2a + (n-1)d)\] QED

Proven as required.

1 mark: writing sum forwards. 1 mark: writing sum reversed. 1 mark: adding and simplifying. 1 mark: concluding with correct formula.

How did you go? (out of 4)

Question 125 marks

A terraced garden has levels numbered 1, 2, 3, ... Level $n$ has area $A_n = 160 \times 0.8^{n-1}$ square metres.

a. Find the area of level 4. [1 mark]

b. Show that the total area of the first $n$ levels is $S_n = 800(1 - 0.8^n)$. [2 marks]

c. Find the maximum possible total area if the garden has infinitely many levels. [1 mark]

d. How many levels are needed for the total area to exceed 95% of the maximum? [1 mark]

a. [1 mark]

$A_4 = 160 \times 0.8^3 = 160 \times 0.512 = 81.92$

81.92 m²

b. [2 marks]

$S_n = \frac{160(1 - 0.8^n)}{1 - 0.8} = \frac{160(1 - 0.8^n)}{0.2} = 800(1 - 0.8^n)$

$S_n = 800(1 - 0.8^n)$ as required. QED

1 mark for correct formula substitution. 1 mark for algebraic simplification to required form.

c. [1 mark]

As $n \to \infty$, $0.8^n \to 0$, so $S_\infty = 800(1 - 0) = 800$

800 m²

d. [1 mark]

$800(1 - 0.8^n) > 0.95 \times 800 \Rightarrow 1 - 0.8^n > 0.95 \Rightarrow 0.8^n < 0.05$. Taking logarithms: $n \log(0.8) < \log(0.05) \Rightarrow n > \frac{\log(0.05)}{\log(0.8)} = 13.43...$

14 levels

How did you go? (out of 5)

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Specialist Mathematics Unit 1

Part A: Beach Monitoring (Arithmetic Sequences & Series)

Part B: Revegetation (Geometric Sequences & Series)

Part C: Project Finances

Part D: Connecting Ideas

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