Mathematics Advanced • Year 12 • Module 7 • Lesson 13
Recurrence Relations for Loans
Practise HSC-style writing on loan recurrences and an extended response on amortisation analysis.
1. Short-answer questions
1.1 A $250,000 home loan is taken at 6% p.a. compounded monthly with monthly repayments of $1,500.
(a) Write the recurrence relation An+1 in terms of An.
(b) Find A1 and A2 to the nearest cent. 3 marks Band 3
1.2 A $30,000 car loan is at 7.2% p.a. compounded monthly with M = $600.
(a) Find the interest I1 charged in Month 1 and the principal reduction P1.
(b) Explain in one sentence why early repayments are mostly interest. 3 marks Band 3-4
1.3 A $20,000 personal loan satisfies the recurrence An+1 = 1.004An − 800 with A0 = $20,000.
(a) Find A1 and A2.
(b) Find the total interest paid in the first 2 months.
(c) State the annual rate (p.a. compounded monthly) implied by the coefficient 1.004. 4 marks Band 4
2. Extended response
2.1 A $400,000 home loan is taken at 5% p.a. compounded monthly. The borrower agrees to monthly repayments of $2,500.
(a) State the monthly rate r and write the recurrence relation An+1 in terms of An.
(b) Build a 3-month amortisation table showing Opening balance | Interest | Principal reduction | Closing balance.
(c) Compute the total interest paid in those 3 months and express it as a percentage of the total $7,500 repaid.
(d) Explain in 2-3 sentences why the principal reduction grows from month to month even though M is fixed, and why this trend accelerates as the loan matures. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct r = 0.05 ÷ 12 = 0.004167 (or exact 5/1200) and correct recurrence An+1 = 1.004167An − 2,500.
Part (b) — 3 marks
• 1 mark — correct opening, interest, repayment, principal, closing for Month 1.
• 1 mark — correct Month 2 row using A1 from Month 1.
• 1 mark — correct Month 3 row using A2 from Month 2.
Part (c) — 1 mark
• 1 mark — correct total interest (≈$4,990) and correct percentage of $7,500 (≈66.5%).
Part (d) — 2 marks
• 1 mark — identifies that the falling balance reduces interest each month, leaving more of M for principal.
• 1 mark — notes that the trend accelerates because the principal reduction itself reduces next month's interest more, creating a compounding loop in the borrower's favour late in the loan.
Your response:
Stuck on (d)? Reference the lesson's "Misconceptions" callout: early repayments are mostly interest because the balance is high.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $250,000 at 6%, M = $1,500 (3 marks)
Sample response. (a) r = 0.06 ÷ 12 = 0.005, so An+1 = 1.005An − 1,500. (b) 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.
Marking notes. 1 mark — correct recurrence. 1 mark — correct A1. 1 mark — correct A2. Watch for sign errors (writing + 1,500 instead of − 1,500) — that's a fundamental misunderstanding and costs the whole mark.
1.2 — $30,000 car loan at 7.2%, M = $600 (3 marks)
Sample response. (a) r = 0.072 ÷ 12 = 0.006. I1 = 30,000 × 0.006 = $180.00. P1 = 600 − 180 = $420.00. (b) The interest portion of any repayment is r × An, and An is largest at the start of the loan, so the interest portion is largest at the start — leaving only a small slice of the repayment to attack principal.
Marking notes. 1 mark — correct I1. 1 mark — correct P1. 1 mark — explanation links high opening balance to high interest portion.
1.3 — Recurrence An+1 = 1.004An − 800, A0 = 20,000 (4 marks)
Sample response. (a) 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. (b) Total interest = I1 + I2 = 20,000 × 0.004 + 19,280 × 0.004 = 80 + 77.12 = $157.12. (c) Coefficient 1.004 implies r = 0.004 per month → annual = 0.004 × 12 = 4.8% p.a. compounded monthly.
Marking notes. 1 mark — correct A1. 1 mark — correct A2. 1 mark — correct total interest computed from the I values. 1 mark — correct annual rate identification.
2.1 — $400,000 / 5% / $2,500 amortisation (7 marks): sample Band-6 response
Sample Band-6 response.
(a) Rate and recurrence. r = 0.05 ÷ 12 = 0.004167 (approx, exact = 1/240). Recurrence: An+1 = 1.004167 × An − 2,500. [1 mark]
(b) Three-month amortisation table.
Month 1: Open $400,000.00 | I = $1,666.67 | P = $833.33 | Close $399,166.67
Month 2: Open $399,166.67 | I = $1,663.20 | P = $836.80 | Close $398,329.87
Month 3: Open $398,329.87 | I = $1,659.71 | P = $840.29 | Close $397,489.58
[3 marks: each row 1 mark]
(c) Interest share over 3 months. Total interest = 1,666.67 + 1,663.20 + 1,659.71 = $4,989.58. Total repaid = $7,500. Interest share = 4,989.58 ÷ 7,500 ≈ 66.5%. [1 mark]
(d) Why principal reduction grows and accelerates. Each month the closing balance is slightly smaller than the opening balance, so next month's interest charge (r × An) is also slightly smaller. With M fixed at $2,500, the smaller interest portion leaves a larger slice of the repayment to attack principal. This is a self-reinforcing loop: a bigger principal cut reduces next month's interest further, which makes the next principal cut even bigger. The effect is small in the first few months (cents) but accelerates over years — by the final years of the loan, almost the entire $2,500 is principal. [2 marks]
Total: 7/7.
Band descriptors for marker.
Band 3: Correct recurrence but table has arithmetic errors (only 1 row fully correct). Total interest computed but no insight in (d). ≈ 3 marks.
Band 4: Correct recurrence, two of three table rows correct, total interest correct, (d) restates "early repayments are mostly interest" without showing the compounding loop. ≈ 4-5 marks.
Band 5: All three rows correct, interest share correct, (d) identifies that falling balance reduces interest each month. ≈ 5-6 marks.
Band 6: Full amortisation table to the cent, correct percentage, and (d) describes the self-reinforcing loop and notes the acceleration is small early but pronounced late in the loan. 7/7.