The specification gives three shapes of relation as examples:
| Form | What it does | Example sequence |
|---|---|---|
| $x_{n+1} = x_n + a$ | Adds a constant each time — an arithmetic sequence | $a = 3$, $x_1 = 2$: $2, 5, 8, 11, \ldots$ |
| $x_{n+1} = a\,x_n$ | Multiplies by a constant — a geometric sequence | $a = 2$, $x_1 = 3$: $3, 6, 12, 24, \ldots$ |
| $x_{n+2} = x_{n+1} + x_n$ | Adds the two previous terms — needs two starting values | $1, 1, 2, 3, 5, 8, \ldots$ (Fibonacci) |
- Write down the starting value.
- Substitute it into the right-hand side to get the next term.
- Repeat, using the term you just found.
- Keep full accuracy throughout; round only the final answer.
A sequence is defined by $x_{n+1} = 2x_n - 3$ with $x_1 = 5$. Find the first five terms.
2 × ANS − 3 and press $=$ repeatedly. Each press gives the next term.$x_{n+2} = x_{n+1} + x_n$ with $x_1 = 2$ and $x_2 = 5$. Find $x_3$, $x_4$ and $x_5$.
For $x_{n+1} = 3x_n + 1$, the fourth term is $x_4 = 40$. Find $x_1$.
For $x_{n+1} = a\,x_n + b$: $L = aL + b$, so $L = \dfrac{b}{1-a}$ (provided $a \neq 1$)
$|a| > 1$: the sequence diverges away from $L$
$x_{n+1} = 0.5x_n + 6$ with $x_1 = 2$. Find the limit and verify it numerically.
$x_{n+1} = 3x_n - 4$ with $x_1 = 3$. Describe the long-term behaviour.
AL11 asks you to use recurrence relationships in modelling, and names compound interest as the example.
$r$ = growth multiplier (e.g. $1.04$ for $4\%$) · $p$ = amount added each period
| Situation | Multiplier $r$ |
|---|---|
| $5\%$ interest gained | $1.05$ |
| $3\%$ lost each year | $0.97$ |
| Value halves each period | $0.5$ |
| Population grows $12\%$ | $1.12$ |
£$1000$ is invested at $5\%$ per year. At the end of each year, after interest, a further £$200$ is deposited. Model this and find the balance after $4$ years.
A patient takes a $20$ mg dose of a drug every day. Each day the body removes $40\%$ of whatever is present. Find the long-term amount in the body just after a dose.
Notation
$x_{n+1}$ is the next term after $x_n$.
Starting value
Essential — the relation alone defines nothing.
Arithmetic
$x_{n+1} = x_n + a$.
Geometric
$x_{n+1} = a\,x_n$.
Fibonacci type
$x_{n+2} = x_{n+1} + x_n$ needs two starts.
Fixed point
Set $x_{n+1} = x_n = L$ and solve.
Converges
When $|a| < 1$.
Diverges
When $|a| > 1$.
Interest
$A_{n+1} = r A_n + p$; $r = 1 + \tfrac{\text{rate}}{100}$.
Accuracy
Use the ANS key; round only at the end.
Write down the first four terms of $x_{n+1} = x_n + 4$ with $x_1 = 3$.
▶ Show solution
$3,\; 7,\; 11,\; 15$
This is an arithmetic sequence with common difference $4$.
Find the first four terms of $x_{n+1} = 3x_n$ with $x_1 = 2$.
▶ Show solution
$2,\; 6,\; 18,\; 54$
A geometric sequence with common ratio $3$.
A sequence has $x_{n+2} = x_{n+1} + 2x_n$, $x_1 = 1$, $x_2 = 3$. Find $x_3$, $x_4$ and $x_5$.
▶ Show solution
$x_3 = 3 + 2(1) = 5$
$x_4 = 5 + 2(3) = 11$
$x_5 = 11 + 2(5) = 21$
Write a recurrence relationship for the sequence $80, 40, 20, 10, \ldots$
▶ Show solution
Each term is half the one before.
$x_{n+1} = 0.5\,x_n$, with $x_1 = 80$.
Find the limit of the sequence given by $x_{n+1} = 0.2x_n + 8$.
▶ Show solution
$L = 0.2L + 8$
$0.8L = 8$, so $L = 10$.
Since $|0.2| < 1$, the sequence converges to $10$ whatever the starting value.
For $x_{n+1} = 4x_n - 6$, the third term is $x_3 = 26$. Find $x_1$.
▶ Show solution
Rearranging, $x_n = \dfrac{x_{n+1}+6}{4}$.
$x_2 = \dfrac{26+6}{4} = 8$
$x_1 = \dfrac{8+6}{4} = 3.5$
Check: $3.5 \to 8 \to 26$ ✓
£$5000$ is invested at $3\%$ per year with no further deposits. Write a recurrence relationship and find the value after $3$ years.
▶ Show solution
$A_{n+1} = 1.03\,A_n$, with $A_0 = 5000$.
$A_1 = 5150$, $A_2 = 5304.50$, $A_3 = 5463.635$
£$5463.64$ to the nearest penny.
Or directly: $5000 \times 1.03^3 = 5463.635$ ✓
State, with a reason, whether $x_{n+1} = 1.5x_n - 2$ with $x_1 = 10$ converges or diverges.
▶ Show solution
Fixed point: $L = 1.5L - 2$, so $0.5L = 2$ and $L = 4$.
The multiplier is $1.5$, and $|1.5| > 1$, so the sequence diverges.
Starting at $10$, which is above $4$, the terms increase without limit: $10, 13, 17.5, 24.25, \ldots$
A lake holds $1200$ fish. Each year $25\%$ are caught, and then $200$ young fish are added. Find the long-term population.
▶ Show solution
$75\%$ remain, so $F_{n+1} = 0.75\,F_n + 200$, with $F_0 = 1200$.
Fixed point: $L = 0.75L + 200 \Rightarrow 0.25L = 200 \Rightarrow L = 800$.
Since $|0.75| < 1$ the population converges to $800$ fish.
It falls towards this from $1200$: $F_1 = 1100$, $F_2 = 1025$, $F_3 = 968.75$, …
A loan of £$8000$ charges $1\%$ interest per month. The borrower repays £$300$ at the end of each month, after the interest is added.
(a) Write a recurrence relationship for the balance $B_n$ after $n$ months. (b) Find the balance after $3$ months. (c) Find the fixed point and explain what it means here. (d) What monthly repayment would leave the balance unchanged forever, and what does that tell the borrower?
▶ Show solution
(a) Interest multiplies the balance by $1.01$, then $300$ is subtracted:
$B_{n+1} = 1.01\,B_n - 300$, with $B_0 = 8000$.
(b) $B_1 = 1.01(8000) - 300 = 8080 - 300 = 7780$
$B_2 = 1.01(7780) - 300 = 7857.80 - 300 = 7557.80$
$B_3 = 1.01(7557.80) - 300 = 7633.378 - 300 = 7333.38$ (to the nearest penny)
(c) $L = 1.01L - 300 \Rightarrow -0.01L = -300 \Rightarrow L = 30\,000$.
The multiplier $1.01$ has $|1.01| > 1$, so this fixed point is unstable: the balance moves away from £$30\,000$. Starting below it at £$8000$, the balance falls — which is what the borrower wants. The loan will be paid off.
Had the starting debt been above £$30\,000$, the same £$300$ payment would have caused the debt to grow without limit.
(d) For no change, the repayment must exactly equal the interest charged on £$8000$:
$0.01 \times 8000 = £80$ per month.
So any repayment above £$80$ reduces the debt; anything below £$80$ and the debt grows forever. The £$300$ payment is comfortably above this threshold, and the £$220$ excess is what actually pays down the loan.