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.