Recurrence Relations for Loans
When you take out a home loan, the bank does not guess your repayments — it calculates them using the same recurrence relation you are about to learn. Every repayment chips away at the principal while interest continues to accrue on what remains.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A $400,000 home loan at 5% p.a. compounded monthly requires monthly repayments of $2,500.
Month 1: Interest $= \$400{,}000 \times 0.05/12 = \$1{,}667$. Principal reduction $= \$2{,}500 - \$1{,}667 = \$833$.
Without calculating — will the principal reduction in Month 2 be larger, smaller, or the same as Month 1? Explain your reasoning.
Key facts
- The loan recurrence relation
- How to calculate interest and principal reduction
- Why early repayments are mostly interest
Concepts
- The amortisation process
- How balance, interest, and principal change over time
- The symmetry between investment and loan recurrences
Skills
- Build loan amortisation tables
- Calculate total interest over a loan's life
- Find when a loan is half paid off
- Compare loans with different rates and terms
For a loan, the recurrence subtracts your repayment instead of adding a deposit. Compare the two:
$A_n$ = outstanding balance at period $n$ | $r$ = interest rate per period | $M$ = regular repayment
This says: Next balance = Current balance with interest charged, minus repayment made.
Example: $A_0 = \$400{,}000$, $r = 0.05/12 \approx 0.004167$, $M = \$2{,}500$.
$A_1 = 1.004167 \times 400{,}000 - 2{,}500 = \$398{,}333.33$
Interest $= 400{,}000 \times 0.004167 = \$1{,}666.67$
Principal reduction $= 2{,}500 - 1{,}666.67 = \$833.33$
Loan recurrence: $A_{n+1} = (1+r)A_n - M$; Interest in period $n$: $I_n = r \times A_n$
Pause — copy the loan recurrence $A_{n+1} = (1+r)A_n - M$ and the period interest formula $I_n = r \times A_n$ — showing that interest is calculated on the outstanding balance, not the original principal — into your book.
Quick check: A loan has $A_0 = \$20{,}000$, $r = 0.005$ per month, $M = \$400$. Which expression gives $A_1$?
We just saw the recurrence $A_{n+1} = (1+r)A_n - M$ and why early repayments are mostly interest (because $A_n$ is large). That raises a question: over the full life of a loan, how does the interest-vs-principal split evolve — and why does a 30-year term cost $140,000 more than a 20-year term on the same principal? This card answers it → amortisation: early repayments are mostly interest (high balance); later repayments are mostly principal (low balance) — extending the term means more high-balance periods paying interest.
Amortisation is the gradual extinction of a loan through regular repayments. The word comes from the Latin ad mortem — "toward death." Early in the loan, interest dominates because the balance is high. Later, principal dominates because the balance is low.
The term makes an enormous difference. Here is $400,000 at 5% p.a.:
| Term | Monthly repayment | Total interest |
|---|---|---|
| 20 years | $2,640 | $233,600 |
| 25 years | $2,340 | $302,000 |
| 30 years | $2,150 | $374,000 |
The 30-year loan "saves" $490/month but costs an extra $140,400 in interest compared to the 20-year loan. The bank loves 30-year loans.
Amortisation = gradual extinction of a loan through regular repayments; Early repayments: mostly interest (high balance). Later repayments: mostly principal (low balance)
Pause — copy the amortisation principle: early repayments = mostly interest (high balance); later repayments = mostly principal (low balance); on a 30-year loan the balance stays above 50% until approximately year 22–23 — into your book.
Did you get this? True or false: on a 30-year home loan, the outstanding balance reaches 50% of the original principal at approximately the 15-year midpoint.
Worked examples · step through 3 in a row
A car loan of $30,000 at 7.2% p.a. compounded monthly. Monthly repayments are $600. Write the recurrence relation and find $A_3$.
$A_2 = 1.006(29{,}580) - 600 = \$29{,}157.48$
$A_3 = 1.006(29{,}157.48) - 600 = \$28{,}732.43$
Using the car loan above ($30,000 at 7.2% p.a., $M = 600$), find the total interest paid in the first 3 months and the interest component of each payment.
$I_3 = 0.006 \times 29{,}157.48 = \$174.94$
A personal loan of $15,000 at 9.6% p.a. compounded monthly. Repayments are $350/month. Find $A_2$ and the total interest paid in the first 2 months.
$A_{n+1} = 1.008A_n - 350$
$A_2 = 1.008(14{,}770) - 350 = \$14{,}538.16$
$I_2 = 0.008 \times 14{,}770 = \$118.16$
Total interest $= \$238.16$
Think & link: A $10,000 loan at 6% p.a. compounded monthly has $M = \$200$/month. What is $A_1$? Type your working and answer.
Common errors · the 3 traps that cost marks
Fill in the blanks. Complete the recurrence for a $50,000 loan at 4.8% p.a. compounded monthly with $M = \$900$/month.
Part 1 — Amortisation table. Set $P = \$400{,}000$, $r = 5\%$ p.a. monthly, $M = \$2{,}500$, term 30 years. Record the balance after 1 year, 5 years, 10 years, and 20 years. At what point is the loan half paid off?
Part 2 — Compare terms. Keep $P = \$400{,}000$, $r = 5\%$. Set $M = \$2{,}640$ (20-year repayment). How many months to pay off? Calculate total repaid and total interest. How much interest is saved vs the 30-year term at $M = \$2{,}500$?
Part 3 — The bank's perspective. A bank offers you a 30-year loan at 4.8% or a 25-year loan at 5.2%, both for $400,000. Which has the lower total interest? Explain why banks advertise low rates with long terms.
Odd one out: Three of these statements about amortisation are true. Which one is FALSE?
Earlier you predicted whether the principal reduction in Month 2 would be larger, smaller, or the same as Month 1.
The answer: Month 2 has $A_1 = \$398{,}333.33$. Interest $= 398{,}333.33 \times 0.004167 = \$1{,}660.42$. Principal reduction $= 2{,}500 - 1{,}660.42 = \$839.58$.
The principal reduction in Month 2 ($839.58) is larger than Month 1 ($833.33) — but only slightly. The balance has dropped, so slightly less interest accrues, meaning slightly more of the repayment hits the principal. This gradual shift is the essence of amortisation. It takes years before the principal reduction becomes substantial.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A mortgage of $250,000 at 6% p.a. compounded monthly has monthly repayments of $1,200. (a) Write the recurrence relation. (b) Find $A_1$. (c) Find $A_2$. (3 marks)
Q2. A car loan of $30,000 at 7.2% p.a. compounded monthly has $M = \$600$/month. (a) Find the interest component of the first repayment. (b) Find the principal reduction in the first repayment. (c) Explain why the principal reduction in the first repayment is so much smaller than the repayment. (3 marks)
Q3. A loan of $20,000 at 4.8% p.a. compounded monthly has $M = \$800$/month. (a) Write the recurrence relation. (b) Find $A_1$ and $A_2$. (c) Calculate the total interest paid in the first 2 months. (4 marks)
Comprehensive answers (click to reveal)
Activity 1: Year 1 ≈ $388,000. Year 5 ≈ $354,000. Year 10 ≈ $302,000. Year 20 ≈ $164,000. Loan drops below $200,000 around year 22–23 — much later than the 15-year midpoint.
Activity 2: At $2,640/month the loan pays off in approximately 240 months (20 years). Total repaid = $2,640 × 240 = $633,600. Total interest = $233,600. 30-year total = $2,500 × 360 = $900,000; interest = $500,000. Interest saved ≈ $266,400.
Activity 3: 30-year at 4.8%: total interest ≈ $350,000. 25-year at 5.2%: total interest ≈ $315,000. Higher-rate shorter term costs less total interest. Banks advertise low rates with long terms because interest compounds over more periods — they earn more overall.
Q1 (3 marks): (a) $A_{n+1} = 1.005A_n - 1{,}200$ [1]. (b) $A_1 = 1.005(250{,}000) - 1{,}200 = \$249{,}050$ [1]. (c) $A_2 = 1.005(249{,}050) - 1{,}200 = \$249{,}095.25$ [1].
Q2 (3 marks): (a) $I_1 = 0.006 \times 30{,}000 = \$180$ [1]. (b) $P_1 = 600 - 180 = \$420$ [1]. (c) The balance is large ($30,000) so the interest charge ($180) is large relative to the repayment ($600) — 30% of the repayment is interest in month 1 [1].
Q3 (4 marks): (a) $r = 0.004$; $A_{n+1} = 1.004A_n - 800$ [1]. (b) $A_1 = 1.004(20{,}000) - 800 = \$19{,}280$ [1]; $A_2 = 1.004(19{,}280) - 800 = \$18{,}557.12$ [1]. (c) $I_1 = 80$, $I_2 = 77.12$; total $= \$157.12$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering loan recurrence questions. Pool: lessons 1–13.
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