Use first-order linear recurrence relations to generate sequences and recognise the patterns of growth and decay they produce.

What this dot point is asking

You must read and use a recurrence, generate terms, and recognise the growth or decay pattern it produces.

What a recurrence relation is

A recurrence relation gives a rule for the next term in terms of the current term, plus an initial value to start. Without the starting term, the rule alone does not pin down the sequence.

This single form underlies the whole finance topic: compound interest ($d = 0$), depreciation, loans ($d$ negative) and savings ($d$ positive).

Generating terms

To generate a sequence, substitute the current term into the rule to get the next, repeating from the starting value. This is exactly what a calculator does when you store a value and apply the rule repeatedly.

Finding the equilibrium

When $0 < R < 1$, the sequence settles to a long-run value where the next term equals the current term. Set $t = R t + d$ and solve: $t(1 - R) = d$, so $t = \dfrac{d}{1 - R}$. This is the same idea as a transition-matrix steady state and as the perpetuity balance in finance.

Translating a worded situation into a recurrence

Most SCSA recurrence questions describe a real process in words, and the skill is turning the words into $t_{n+1} = R t_n + d$. Identify two things: the multiplier and the regular change. A percentage change gives the multiplier ("falls by $10\%$" gives $R = 0.9$, "rises by $5\%$" gives $R = 1.05$); a fixed amount added or removed each period gives $d$ (a $\$300$ deposit gives $d = +300$, a $\$300$ withdrawal gives $d = -300$). Then record the starting value as $t_0$.

For example, "a tank holds $2000$ litres; each day $5\%$ evaporates and $80$ litres is added" becomes $V_{n+1} = 0.95 V_n + 80$, with $V_0 = 2000$. Because $0 < R = 0.95 < 1$, the volume approaches the equilibrium $V = \dfrac{80}{1 - 0.95} = \dfrac{80}{0.05} = 1600$ litres in the long run.

Linking to the financial models

This single recurrence underlies the whole Unit 3 and Unit 4 finance strand. Compound interest sets $d = 0$ and $R > 1$; reducing-balance depreciation sets $d = 0$ and $0 < R < 1$; a reducing-balance loan keeps $R > 1$ with $d$ negative (a repayment); a savings annuity keeps $R > 1$ with $d$ positive (a deposit). Recognising which case a question describes tells you immediately how the sequence will behave and which long-run outcome (growth, decay, or equilibrium) to expect.