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What even is this? Instead of a formula that jumps straight to any term (like tₙ = a + (n−1)d), a recurrence relation says: "to get the next term, do this to the current term." You build the sequence step by step. It looks more complicated, but it's just a chain, each term unlocks the next one.
Section 1 · The General Form
tn+1 = r × tn + d
with a starting value t₁ always given alongside the relation
tₙ₊₁
The next term
The term you're calculating, one position ahead of where you are now.
r
Multiplier
Multiply the current term by r first. If r=1, nothing gets multiplied, you're just adding.
d
Fixed amount added
Added (or subtracted if negative) after the multiplication. Could be a deposit, repayment, or 0.
Each term feeds directly into the next. You always need the starting value t₁, the recurrence alone isn't enough.
Section 2 · Three Types to Recognise
| If... | Type | What it means | Example |
|---|---|---|---|
| r = 1, d ≠ 0 | Arithmetic | No multiplying, just adding the same amount each time. Same as tₙ = a + (n−1)d. | tₙ₊₁ = tₙ + 5, t₁ = 3 → 3, 8, 13, 18… |
| r ≠ 1, d = 0 | Geometric | Multiply by the same ratio each time, nothing added. Same as tₙ = arⁿ⁻¹. | tₙ₊₁ = 3·tₙ, t₁ = 2 → 2, 6, 18, 54… |
| r ≠ 1, d ≠ 0 | Financial | Both multiplying AND adding. This is the one that models loans, annuities, and depreciation. | tₙ₊₁ = 1.1·tₙ + 200 → compound growth with regular deposits |
Section 3 · Generating Terms Step by Step
There's no shortcut formula here, just substitute the current term in, calculate the next one, repeat. The key is being methodical.
Example: t₁ = 200, tₙ₊₁ = 2·tₙ − 100 (find the first 4 terms)
1
t₁ = 200 (given)
2
t₂ = 2 × t₁ − 100 = 2 × 200 − 100 = 400 − 100 = 300
3
t₃ = 2 × t₂ − 100 = 2 × 300 − 100 = 600 − 100 = 500
4
t₄ = 2 × t₃ − 100 = 2 × 500 − 100 = 1000 − 100 = 900
Section 4 · Writing a Recurrence from Context
The exam will describe a financial situation and ask you to write the recurrence relation. Identify two things: what gets multiplied (that's r) and what gets added or subtracted (that's d).
Compound Interest (no deposits)
"Starts at $5000. Grows by 8% each year."
tₙ₊₁ = 1.08 × tₙ, t₁ = 5000
Investment with deposits
"Starts at $1000. Grows by 5% per year, then $200 added."
tₙ₊₁ = 1.05 × tₙ + 200, t₁ = 1000
Depreciation (flat rate)
"A machine worth $20,000 loses $2,000 in value each year."
tₙ₊₁ = tₙ − 2000, t₁ = 20000
Reducing balance loan
"Loan of $1000. 20% interest added, then $300 repaid each period."
tₙ₊₁ = 1.2 × tₙ − 300, t₁ = 1000
⚠️ Watch the sign on d: deposits and additions → d is positive. Repayments and depreciation losses → d is negative. Getting this backwards is a very common exam mistake.
Section 5 · Worked Example
☀️ Maria's Solar Investment
Maria invests $2000 in a solar energy fund. Each year it earns 5% interest, and she tops it up with an extra $100. She wants to track the value over the first 4 years.
Step 1 · Write the recurrence relation
Identify r and d, then write the relation.
Grows by 5% → multiply by 1.05 → r = 1.05
Adds $100 each year → d = +100
Starting value: t₁ = 2000
tₙ₊₁ = 1.05 × tₙ + 100, t₁ = 2000
Adds $100 each year → d = +100
Starting value: t₁ = 2000
tₙ₊₁ = 1.05 × tₙ + 100, t₁ = 2000
Step 2 · Generate the first 4 terms
Calculate the value of the investment at the end of each year.
| Year | Value (tₙ) | Calculation | New value (tₙ₊₁) |
|---|---|---|---|
| 1 | $2000 | 1.05 × 2000 + 100 | $2200 |
| 2 | $2200 | 1.05 × 2200 + 100 | $2410 |
| 3 | $2410 | 1.05 × 2410 + 100 | $2630.50 |
| 4 | $2630.50 | , | ✓ Value at end of year 4 |
Step 3 · Interpret
What type of sequence is this, and why is the growth speeding up?
r = 1.05 ≠ 1 and d = 100 ≠ 0 → financial sequence (neither pure arithmetic nor geometric).
The growth speeds up because each year's interest is calculated on a slightly larger balance, the interest compounds, even though she's also depositing a fixed amount.
The growth speeds up because each year's interest is calculated on a slightly larger balance, the interest compounds, even though she's also depositing a fixed amount.
Section 6 · Practice Questions
Tap to reveal the worked answer.
Question 1
Given: tₙ₊₁ = 1.3·tₙ − 50, t₁ = 200.
What is t₂?
What is t₂?
▼
t₂ = 1.3 × t₁ − 50
t₂ = 1.3 × 200 − 50
t₂ = 260 − 50 = 210
t₂ = 1.3 × 200 − 50
t₂ = 260 − 50 = 210
✓ t₂ = 210
Question 2, Arithmetic type
A sequence is defined by tₙ₊₁ = tₙ + 8, t₁ = 15. Find t₅.
▼
This is arithmetic (r = 1, d = 8).
t₂ = 15 + 8 = 23
t₃ = 23 + 8 = 31
t₄ = 31 + 8 = 39
t₅ = 39 + 8 = 47
(You could also use tₙ = 15 + (n−1)×8. t₅ = 15 + 32 = 47 ✓)
t₂ = 15 + 8 = 23
t₃ = 23 + 8 = 31
t₄ = 31 + 8 = 39
t₅ = 39 + 8 = 47
(You could also use tₙ = 15 + (n−1)×8. t₅ = 15 + 32 = 47 ✓)
✓ t₅ = 47
Question 3, Geometric type
A sequence is defined by tₙ₊₁ = 4·tₙ, t₁ = 3. Find t₄.
▼
This is geometric (r = 4, d = 0).
t₂ = 4 × 3 = 12
t₃ = 4 × 12 = 48
t₄ = 4 × 48 = 192
(Also: t₄ = 3 × 4³ = 3 × 64 = 192 ✓)
t₂ = 4 × 3 = 12
t₃ = 4 × 12 = 48
t₄ = 4 × 48 = 192
(Also: t₄ = 3 × 4³ = 3 × 64 = 192 ✓)
✓ t₄ = 192
Question 4, Financial sequence
A loan: tₙ₊₁ = 1.2·tₙ − 300, t₁ = 1000. What is the outstanding balance t₄?
▼
t₂ = 1.2 × 1000 − 300 = 1200 − 300 = 900
t₃ = 1.2 × 900 − 300 = 1080 − 300 = 780
t₄ = 1.2 × 780 − 300 = 936 − 300 = 636
t₃ = 1.2 × 900 − 300 = 1080 − 300 = 780
t₄ = 1.2 × 780 − 300 = 936 − 300 = 636
✓ t₄ = $636 still owed
Question 5, Write the recurrence
A fish tank starts with 50 fish. Each week, 20% of the fish die, and 15 new fish are added. Write the recurrence relation and find t₃.
▼
20% die → 80% survive → r = 0.8
15 added → d = +15
Recurrence: tₙ₊₁ = 0.8·tₙ + 15, t₁ = 50
t₂ = 0.8 × 50 + 15 = 40 + 15 = 55
t₃ = 0.8 × 55 + 15 = 44 + 15 = 59
15 added → d = +15
Recurrence: tₙ₊₁ = 0.8·tₙ + 15, t₁ = 50
t₂ = 0.8 × 50 + 15 = 40 + 15 = 55
t₃ = 0.8 × 55 + 15 = 44 + 15 = 59
✓ tₙ₊₁ = 0.8·tₙ + 15, t₁ = 50. After 3 weeks: 59 fish.
The Score Tracker, Escape Room
Josh's strategy game tournament closes at midnight. Six rounds of recurrence relation challenges, arithmetic, geometric, and financial sequences, before the leaderboard locks.
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