Mathematics Advanced • Year 12 • Module 7 • Lesson 13

Recurrence Relations for Loans

Build fluency iterating the loan recurrence A_n+1 = (1+r)A_n − M, and computing interest and principal components.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the loan recurrence (with repayment M each period):

A_n+1 = ____________________

Q1.2 Interest in period n on outstanding balance A_n:

I_n = ____________________

Q1.3 A monthly repayment is $2,500. The balance is $400,000 and the monthly rate is 0.004167. State the interest component and the principal reduction for this month.

Stuck? Revisit lesson § Formula Reference and § Loan Recurrence.

2. Worked example — Iterate a car loan

Follow every line. Each step has a short reason.

Problem. $30,000 car loan at 7.2% p.a. compounded monthly. Monthly repayments $600. Find A₃ and total interest in the first 3 months.

Step 1 — Compute the monthly rate.

r = 0.072 ÷ 12 = 0.006 per month

Step 2 — Write the recurrence.

A_n+1 = 1.006 × A_n − 600

Step 3 — Iterate, tracking interest each period.

A₀ = 30,000.00

I₁ = 30,000 × 0.006 = $180.00 → A₁ = 30,000 + 180 − 600 = $29,580.00

I₂ = 29,580 × 0.006 = $177.48 → A₂ = 29,580 + 177.48 − 600 = $29,157.48

I₃ = 29,157.48 × 0.006 = $174.94 → A₃ = 29,157.48 + 174.94 − 600 = $28,732.42

Conclusion. A₃ = $28,732.42. Total interest paid in first 3 months = 180.00 + 177.48 + 174.94 = $532.42 (89% of the principal reduction).

3. Faded example — fill in the missing steps

A personal loan of $15,000 at 9.6% p.a. compounded monthly. Repayments are $350/month. Find A₂ and total interest in the first 2 months. 4 marks

Step 1 — Monthly rate:

r = 0.096 ÷ 12 = ______________

Step 2 — Recurrence:

A_n+1 = ______________ × A_n − ______________

Step 3 — Iterate.

I₁ = 15,000 × ______________ = $______________

A₁ = 15,000 + ______________ − 350 = $______________

I₂ = ______________ × 0.008 = $______________

A₂ = ______________ + ______________ − 350 = $______________

Conclusion. Total interest in first 2 months = $______________.

Stuck? Revisit lesson § Worked Example — Try It Now.

4. Graduated practice

Show the substitution and the final value (to nearest cent unless stated). Use the same time units for r and n.

Foundation — direct substitution (4 questions)

Q	Scenario	Working & answer
4.1 1	Write the loan recurrence with r = 0.005 and M = $1,200.
4.2 1	5.4% p.a. compounded monthly — state r per month.
4.3 1	A₀ = $250,000 at r = 0.005, repayment M = $1,200. State I₁.
4.4 1	For 4.3, state the Month 1 principal reduction P₁.

Standard — typical HSC difficulty (6 questions)

Show at least one substitution line and one evaluation line.

4.5 $250,000 loan at 6% p.a. compounded monthly, M = $1,500. Find A₁ and A₂. 2 marks

4.6 $20,000 personal loan at 4.8% p.a. compounded monthly, M = $800. Find A₂ and the total interest paid in the first 2 months. 2 marks

4.7 $400,000 home loan at 5% p.a. compounded monthly, M = $2,500. Find I₁, P₁ (principal reduction) and A₁. 2 marks

4.8 A loan satisfies A_n+1 = 1.004A_n − 800, A₀ = $20,000. Find A₃. 2 marks

4.9 $50,000 student loan at 6% p.a. compounded monthly, M = $1,000. Find A₃ and the total interest paid in those 3 months. 2 marks

4.10 A car loan recurrence is A_n+1 = 1.005A_n − 600 with A₀ = $25,000. Find A₂ and explain (1 sentence) whether the principal reduction in Month 2 is larger or smaller than in Month 1, and why. 2 marks

Extension — combine concepts (2 questions)

4.11 For the $400,000 / 5% / M = $2,500 loan, compute the principal reduction in Month 1 and in Month 2. Express each as a percentage of the $2,500 repayment, and state by how many cents the principal reduction increased from Month 1 to Month 2. 3 marks

4.12 A $100,000 loan satisfies A_n+1 = 1.005A_n − M, where M is the borrower's chosen repayment. Find the smallest M (to the nearest dollar) such that A₁ < A₀ — that is, the smallest repayment that actually reduces the balance. Explain what would happen if M were smaller than this value. 3 marks

Stuck on 4.12? Solve 1.005(100,000) − M < 100,000.

5. Self-check the easy 3

Tick the first three once you have checked the method works.

For 4.1 my recurrence is A_n+1 = 1.005A_n − 1,200.

For 4.2 I got r = 0.054 ÷ 12 = 0.0045 per month.

For 4.3 I got I₁ = 250,000 × 0.005 = $1,250.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Loan recurrence

A_n+1 = (1 + r)A_n − M.

Q1.2 — Interest in period n

I_n = r × A_n.

Q1.3 — $400,000 × 0.004167

I = 400,000 × 0.004167 = $1,666.67. Principal reduction = 2,500 − 1,666.67 = $833.33.

Q3 — Faded example: $15,000 at 9.6%, M = $350

r = 0.096 ÷ 12 = 0.008. Recurrence: A_n+1 = 1.008A_n − 350.
I₁ = 15,000 × 0.008 = $120.00. A₁ = 15,000 + 120 − 350 = $14,770.00.
I₂ = 14,770 × 0.008 = $118.16. A₂ = 14,770 + 118.16 − 350 = $14,538.16.
Total interest = 120 + 118.16 = $238.16.

Q4.1 — Loan recurrence with r = 0.005, M = 1,200

A_n+1 = 1.005A_n − 1,200.

Q4.2 — Monthly rate for 5.4% p.a.

r = 0.054 ÷ 12 = 0.0045.

Q4.3 — I₁ on $250,000 at r = 0.005

I₁ = 250,000 × 0.005 = $1,250.00.

Q4.4 — P₁ for M = $1,200

P₁ = M − I₁ = 1,200 − 1,250 = −$50.00. The repayment is less than the interest charged, so the principal does not reduce — the balance actually grows by $50 this month.

Q4.5 — $250,000 at 6% monthly, M = $1,500

r = 0.005. A₁ = 1.005(250,000) − 1,500 = 251,250 − 1,500 = $249,750.00. A₂ = 1.005(249,750) − 1,500 = 250,998.75 − 1,500 = $249,498.75.

Q4.6 — $20,000 at 4.8% monthly, M = $800

r = 0.004. I₁ = 20,000 × 0.004 = $80. A₁ = 20,000 + 80 − 800 = $19,280.00. I₂ = 19,280 × 0.004 = $77.12. A₂ = 19,280 + 77.12 − 800 = $18,557.12. Total interest = 80 + 77.12 = $157.12.

Q4.7 — $400,000 at 5% monthly, M = $2,500

r = 0.05 ÷ 12 = 0.004167. I₁ = 400,000 × 0.004167 = $1,666.67. P₁ = 2,500 − 1,666.67 = $833.33. A₁ = 400,000 − 833.33 = $399,166.67.

Q4.8 — A_n+1 = 1.004A_n − 800, A₀ = 20,000, find A₃

A₁ = 1.004(20,000) − 800 = 20,080 − 800 = $19,280.00.
A₂ = 1.004(19,280) − 800 = 19,357.12 − 800 = $18,557.12.
A₃ = 1.004(18,557.12) − 800 = 18,631.35 − 800 = $17,831.35.

Q4.9 — $50,000 at 6% monthly, M = $1,000, find A₃ and 3-month interest

r = 0.005. I₁ = 50,000 × 0.005 = $250. A₁ = 50,000 + 250 − 1,000 = $49,250.00.
I₂ = 49,250 × 0.005 = $246.25. A₂ = 49,250 + 246.25 − 1,000 = $48,496.25.
I₃ = 48,496.25 × 0.005 = $242.48. A₃ = 48,496.25 + 242.48 − 1,000 = $47,738.73.
Total interest = 250 + 246.25 + 242.48 = $738.73.

Q4.10 — A_n+1 = 1.005A_n − 600, A₀ = 25,000, A₂

A₁ = 1.005(25,000) − 600 = 25,125 − 600 = $24,525.00. A₂ = 1.005(24,525) − 600 = 24,647.63 − 600 = $24,047.63. Principal reduction is larger in Month 2: the balance has fallen from $25,000 to $24,525 so the interest for Month 2 (≈$122.63) is smaller than for Month 1 ($125), leaving more of the $600 to attack principal.

Q4.11 — $400,000 / 5% / M = $2,500: principal in Month 1 vs Month 2

Month 1: I = $1,666.67 → P = $833.33 (33.3% of $2,500).
Month 2: A₁ = $399,166.67. I = 399,166.67 × 0.004167 = $1,663.20 → P = 2,500 − 1,663.20 = $836.80 (33.5% of $2,500).
Principal reduction grew by 836.80 − 833.33 = $3.47. The shift is tiny in Month 2 but accumulates over years — the essence of amortisation.

Q4.12 — Minimum M for A₁ < A₀ on $100,000 at r = 0.005

A₁ = 1.005(100,000) − M = 100,500 − M. For A₁ < A₀ = 100,000, we need 100,500 − M < 100,000, so M > $500. The minimum integer M is $501. If M ≤ $500, the repayment does not even cover the monthly interest of $500, so the balance grows each month and the loan can never be cleared (negative amortisation).