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Module 7 · L13 of 20 ~40 min ⚡ +95 XP available

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.

Today's hook, On a $\$400,000$ home loan at 5%, your first monthly repayment of $\$2,500$ sends only $\$833$ to your actual debt. The other $\$1,667$ disappears to the bank as interest. And on a genuine 30-year loan at this rate, why does it take about 20 years before half the principal is gone?
0/5QUESTS

Get oriented

Recall what you already know and settle the terms for loan recurrences.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Think first, your gut answer
+5 XP warm-up

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.

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What you'll master
Know

Key facts

  • The loan recurrence relation
  • How to calculate interest and principal reduction
  • Why early repayments are mostly interest
Understand

Concepts

  • The amortisation process
  • How balance, interest, and principal change over time
  • The symmetry between investment and loan recurrences
Can do

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
03
Key terms
Recurrence relationA formula that expresses each term using the previous term: $A_{n+1} = (1+r)A_n - M$.
PrincipalThe outstanding amount of the loan at any point in time ($A_n$).
Interest per period$I_n = r \times A_n$, the interest charged on the current balance.
Principal reduction$P_n = M - I_n$, the part of the repayment that actually reduces the debt.
AmortisationThe gradual extinction of a loan through regular repayments over time.
Amortisation tableA period-by-period table showing balance, interest, principal reduction for each payment.

Write the recurrence

Express a loan balance as one step depending on the step before it.

04
The loan recurrence relation
core concept

For a loan, the recurrence subtracts your repayment instead of adding a deposit. Compare the two:

Investment recurrence
$A_{n+1} = (1+r)A_n + D$, balance grows as you add deposits.
Loan recurrence
$A_{n+1} = (1+r)A_n - M$, balance falls as you make repayments.
$$A_{n+1} = (1+r)A_n - M$$

$A_n$ = 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 $= 2{,}500 - 1{,}666.67 = \$833.33$

67% went to interest in month 1
The key insight. In Month 1, only 33% of your $\$2,500$ repayment reduces your actual debt. The bank collects $\$1,667$ before a single dollar touches the principal. This ratio gradually shifts over the life of the loan, but it takes years before principal dominates.

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 $A_1$?

See how a loan dies

Watch the split between interest and principal shift over the life of the loan.

05
Amortisation, how loans die
+5 XP to read

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.

Half-paid-off rule. On a 30-year loan at 5%, the balance does not drop below 50% until about year 20, much later than the 15-year midpoint. This is because most early repayments go to interest rather than principal.

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 about year 20, 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.

Work the examples

Follow three worked recurrence problems end to end.

Worked examples · step through 3 in a row

PROBLEM 1 · WRITE THE RECURRENCE & FIRST BALANCE

A car loan of $\$30,000$ at 7.2% p.a. compounded monthly. Monthly repayments are $\$600$. Write the recurrence relation and find $A_3$.

1
$r = 0.072 \div 12 = 0.006$ per month
Divide the annual rate by 12 to get the monthly rate.
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PROBLEM 2 · INTEREST VS PRINCIPAL

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.

1
$I_1 = 0.006 \times 30{,}000 = \$180$
Interest = rate $\times$ opening balance for that month.
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PROBLEM 3 · PERSONAL LOAN

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.

1
$r = 0.096 \div 12 = 0.008$ per month
$A_{n+1} = 1.008A_n - 350$
Write the recurrence first.
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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.

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Avoid the traps, then revisit

Fix the even-split assumption, run the explorer, then name what has changed.

Common errors · the 3 traps that cost marks

Trap 01
50/50 split from the start
Many students assume early repayments are split evenly between interest and principal. They are not. On a large loan at typical rates, 60–70% goes to interest in the early years.
Trap 02
Lower repayment = better deal
A lower monthly repayment extends the term and dramatically increases total interest. A $\$400$k loan at 5% costs $\$374$k interest over 30 years but only $\$234$k over 20 years. The "saving" of $\$490$/month costs $140,000 extra.
Trap 03
Forgetting to convert the rate
If the loan compounds monthly, divide the annual rate by 12 to get $r$. Using the annual rate directly in the monthly recurrence will massively overstate interest.

Fill in the blanks. Complete the recurrence for a $\$50,000$ loan at 4.8% p.a. compounded monthly with $M = \$900$/month.

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Activity, loan recurrence explorer
+10 XP activity
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Part 1, Amortisation table. Set $P = \$400{,}000$, $r = 5\%$ p.a. monthly and the 30-year repayment $M = \$2{,}150$ from the table above. Record the balance after 1 year, 5 years, 10 years, and 20 years. At what point is the loan half paid off?

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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 its repayment of $M = \$2{,}150$? (Why would comparing against the lesson's $\$2{,}500$ example be wrong here?)

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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?

11
Revisit your thinking

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.

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Practise loan recurrences

Write full short-answer responses, then check them against the model answers.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 43 marks

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)

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ApplyBand 43 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)

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AnalyseBand 54 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)

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Comprehensive answers (click to reveal)

Activity 1: Year 1 ≈ $\$394{,}000$. Year 5 ≈ $\$367{,}000$. Year 10 ≈ $\$325{,}000$. Year 20 ≈ $\$201{,}000$. The balance drops below $\$200{,}000$ at about year 20.2, 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 = $\$233,600$. The 30-year comparison must use the 30-year repayment, not $\$2{,}500$: at $\$2{,}500$ the loan is gone after about 265 months, so 360 payments never happen. 30-year total ≈ $\$2{,}150$ × 359 = $\$771{,}900$; interest ≈ $\$371{,}900$. Interest saved ≈ $\$138{,}300$.

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].