Mathematics Advanced • Year 12 • Module 7 • Lesson 13
Recurrence Relations for Loans
Apply the loan recurrence to amortisation tables, term comparisons and total-interest analyses.
Problem 1 — The $400,000 / 5% / $2,500 home loan (Think First scenario)
A $400,000 home loan at 5% p.a. compounded monthly has monthly repayments of $2,500.
Set up: What are we solving for?
(i) Write the recurrence relation in terms of An. 1 mark
(ii) Find the interest, principal reduction and ending balance for Month 1 and Month 2. 3 marks
(iii) Compare the principal reductions in Month 1 vs Month 2. State whether the second is larger or smaller, by how many cents, and explain in 1 sentence why this is the start of amortisation. 3 marks
Stuck? Revisit lesson § Think First and § Loan Recurrence.Problem 2 — A $30,000 car loan over the first quarter
A $30,000 car loan at 7.2% p.a. compounded monthly. The borrower pays $600 each month.
Set up: What are we solving for?
(i) Build a 3-month amortisation table (Month | Opening balance | Interest | Repayment | Principal reduction | Closing balance). 3 marks
(ii) Find the total interest paid in those 3 months and what percentage of the total repaid ($1,800) was interest. 2 marks
(iii) Explain in 1 sentence why this percentage will fall as the loan progresses, even though M stays at $600. 1 mark
Problem 3 — 20-year vs 30-year repayment on the same $400,000 loan
The bank offers two terms at 5% p.a. compounded monthly: M20 = $2,640/month for 20 years and M30 = $2,150/month for 30 years.
Set up: What are we solving for?
(i) Calculate the total amount repaid under each schedule (M × number of months) and the total interest paid. 3 marks
(ii) Calculate the dollar gap in total interest. Is the 30-year option ever cheaper in total interest? 2 marks
(iii) Explain in 1-2 sentences why the bank prefers to sell 30-year loans. 2 marks
Problem 4 — The repayment-too-low trap
A $250,000 loan at 6% p.a. compounded monthly. The borrower can only afford $1,200/month.
Set up: What are we solving for?
(i) Compute I1 (the interest charged in Month 1) and compare it to M = $1,200. State A1 and the principal reduction in Month 1. 2 marks
(ii) Determine the minimum M (to the nearest dollar) such that the balance actually reduces in Month 1. 2 marks
(iii) Explain in 1-2 sentences what happens to the balance over many months when M is below this threshold. Why do banks call this "negative amortisation"? 2 marks
Stuck? The breakeven is M = rA0.Problem 5 — A rate hike mid-loan
A $200,000 loan at 4.8% p.a. compounded monthly with M = $1,400. After 6 months the rate rises to 5.4% p.a. compounded monthly.
Set up: What are we solving for?
(i) Write the recurrence for months 1–6 and find A6. (Use the closed-form expression for a loan: An = A0(1+r)n − M × [(1+r)n − 1] ÷ r.) 3 marks
(ii) Write the recurrence for months 7 onward (new rate, same M) using A6 as the new starting balance. Iterate one step to find A7. 2 marks
(iii) Explain in 1 sentence why the closed-form formula for the whole loan would be wrong in this scenario. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — $400,000 / 5% / $2,500
Set up. Use the loan recurrence to compute Month 1 and Month 2 and contrast principal reductions.
(i) r = 0.05 ÷ 12 = 0.004167. Recurrence: An+1 = 1.004167 × An − 2,500.
(ii) Month 1: I1 = 400,000 × 0.004167 = $1,666.67. P1 = 2,500 − 1,666.67 = $833.33. A1 = $399,166.67. Month 2: I2 = 399,166.67 × 0.004167 = $1,663.20. P2 = 2,500 − 1,663.20 = $836.80. A2 = $398,329.87.
(iii) Principal reduction grew from $833.33 to $836.80 — a gain of $3.47. Each month the balance is smaller, so less interest is charged, leaving more of the $2,500 to attack principal — that compounding shift is amortisation.
Problem 2 — $30,000 car loan
Set up. Tabulate 3 months and compute interest share.
(i) r = 0.006 per month.
Month 1: Open $30,000.00 | I = $180.00 | M = $600 | P = $420.00 | Close $29,580.00.
Month 2: Open $29,580.00 | I = $177.48 | M = $600 | P = $422.52 | Close $29,157.48.
Month 3: Open $29,157.48 | I = $174.94 | M = $600 | P = $425.06 | Close $28,732.42.
(ii) Total interest = 180.00 + 177.48 + 174.94 = $532.42. Total repaid = $1,800. Interest share = 532.42 ÷ 1,800 ≈ 29.6%.
(iii) The balance keeps falling each month, so the interest charged each month also falls; with M fixed at $600, more of each repayment becomes principal — the interest share of every repayment declines steadily.
Problem 3 — Two terms on $400,000 at 5%
Set up. Compare total repaid and total interest across two terms.
(i) 20-year: total repaid = 2,640 × 240 = $633,600; total interest = 633,600 − 400,000 = $233,600. 30-year: total repaid = 2,150 × 360 = $774,000; total interest = $374,000.
(ii) Gap = 374,000 − 233,600 = $140,400 more interest on the 30-year loan. The 30-year option is never cheaper in total interest at the same rate — longer term always costs more interest.
(iii) Banks earn from interest. A 30-year schedule delivers an extra $140,400 in interest income per $400,000 lent, so they market longer terms (with lower monthly repayments) to make the loan look "more affordable" even though the lifetime cost is higher.
Problem 4 — The repayment-too-low trap
Set up. Compute Month 1 interest, compare to M, then find breakeven M.
(i) r = 0.005. I1 = 250,000 × 0.005 = $1,250.00, which exceeds M = $1,200. A1 = 1.005(250,000) − 1,200 = 251,250 − 1,200 = $250,050.00 — the balance grows by $50. P1 = 1,200 − 1,250 = −$50.00.
(ii) Breakeven: M = rA0 = 250,000 × 0.005 = $1,250. Minimum M (to the nearest dollar) such that the balance reduces in Month 1 = $1,251.
(iii) Each month M < rAn, so the unpaid interest is added to the balance, which raises next month's interest charge even higher. The balance spirals upward — the bank still calls each $1,200 a "repayment", but the borrower is going backwards. This is negative amortisation.
Problem 5 — Rate hike mid-loan
Set up. Project months 1–6 at the old rate, then chain a new recurrence at the new rate.
(i) Phase 1: r = 0.048 ÷ 12 = 0.004. Recurrence An+1 = 1.004An − 1,400. Closed form: (1.004)⁶ = 1.024214. A6 = 200,000(1.024214) − 1,400 × (1.024214 − 1) ÷ 0.004 = 204,842.80 − 1,400 × 6.05350 = 204,842.80 − 8,474.90 ≈ $196,367.90.
(ii) Phase 2: new r = 0.054 ÷ 12 = 0.0045. Recurrence An+1 = 1.0045An − 1,400 with A6 = $196,367.90. A7 = 1.0045(196,367.90) − 1,400 = 197,251.55 − 1,400 ≈ $195,851.55.
(iii) The closed form An = A0(1+r)n − M × [(1+r)n − 1] ÷ r assumes a single constant r for every period. With a rate change at month 7, no single r is valid for the whole loan, so the formula gives the wrong balance — the recurrence (one step at a time, using the current month's r) is the correct tool.