Use first-order linear recurrence relations of the form V(n+1) = R V(n) + d to model and analyse practical situations.

What this dot point is asking

Arithmetic sequences only add, and geometric sequences only multiply. Many real processes do both at once: a savings account earns interest and also receives a regular deposit; a fish population breeds and is also harvested. The first-order linear recurrence captures both actions in one rule.

The two special cases you already know sit inside this one form. When $R = 1$ the rule becomes $V_{n+1} = V_n + d$, an arithmetic sequence. When $d = 0$ it becomes $V_{n+1} = R\,V_n$, a geometric sequence.

Generating terms

You usually build the sequence step by step on a calculator: start at $V_0$, multiply by $R$, add $d$, and repeat. The order matters: multiply first, then add or subtract the constant.

Long-term behaviour

What happens after many steps depends on $R$. If $R = 1$, the sequence grows or shrinks by a constant $d$ forever (linear). If $R > 1$, the multiplying effect eventually dominates and the terms grow without bound. If $0 < R < 1$, the terms settle toward a steady value called the equilibrium or steady state.

Choosing the model

Use a plain geometric model when nothing is added or removed beyond the percentage change. Use the full first-order linear recurrence whenever there is also a fixed regular amount, such as a repayment, deposit, or harvest, layered on top of the percentage change.

Converting between recursive and explicit forms

TASC questions often give a rule in explicit (closed) form and ask for the recursive form, or the reverse. The link is direct. An explicit arithmetic rule $t_n = a + (n-1)d$ comes from the recurrence $t_{n+1} = t_n + d$ with $t_1 = a$. An explicit geometric rule $t_n = a r^{\,n-1}$ comes from $t_{n+1} = r\,t_n$ with $t_1 = a$. To go from explicit to recursive, read off the multiplier (the base of the power, which becomes $R$) and the constant added each step (which becomes $d$), then state the starting term. To go the other way, identify $R$ and $d$ and recognise the matching closed form, or simply iterate.

Reading the steady state from a graph

When a first-order linear recurrence with $0 < R < 1$ is plotted, the points approach a horizontal line at the equilibrium value $V = \dfrac{d}{1 - R}$. If the start is above this line the sequence falls toward it; if below, it rises toward it. The equilibrium does not depend on the starting value, only on $R$ and $d$, which is why two funds with the same rate and the same regular amount end up at the same long-run level no matter where they begin.

A complete answer writes the recurrence in the form $V_{n+1} = R\,V_n + d$ with the correct $R$ and signed $d$, states $V_0$, generates the requested terms in the right order, converts between recursive and explicit forms when asked, and finds the steady state from $V = d/(1 - R)$.