Modelling linear growth
Learning Objectives
- Recognise situations whose underlying structure is an arithmetic progression.
- Translate a verbal description into the parameters \(a_1\), \(d\), and \(n\).
- Apply the general-term and sum formulas to answer questions about individual values, totals, and threshold crossings.
- Identify the limits of the linear model and motivate the need for geometric progressions in the next unit.
Motivation
A great many everyday quantities change by the same amount over equal intervals. A monthly savings deposit, a training schedule that adds the same distance each day, a piece of equipment that loses a fixed amount of value every year — each of these is an arithmetic progression in disguise. The work in this lesson is to learn how to see that disguise.
Once we recognise the structure, the formulas of the previous two lessons do the rest of the work. The \(n\)-th term tells us the value at any given step; the partial sum tells us the cumulative total over the first \(n\) steps. Together, they answer the typical questions one wants to ask: what will the value be after \(n\) periods? what is the total accumulated by then? after how many steps does the value cross some threshold?
A modelling protocol
The translation from words to mathematics, when an arithmetic progression is the right tool, follows a short checklist:
Identify the changing quantity. What does \(a_n\) represent? — a balance, a distance, a value.
Identify the step. What event takes us from \(a_n\) to \(a_{n+1}\)? — a month passes, a day's training is added, a year of depreciation occurs.
Determine \(a_1\). The value after the first step (or, if more convenient, the initial value, sometimes written \(a_0\)).
Determine \(d\). The fixed change per step — positive for growth, negative for decay.
Verify the structure. Is the change really the same at every step? If it is not, the model is wrong and a different tool is needed.
Translate the question. Are we asked for a specific term (use the general term), for a total (use the partial sum), or for the smallest \(n\) reaching some threshold (solve the inequality)?
Maria's savings
Maria opens a savings account on the first of January with an initial deposit of \(\,€200\). At the end of every subsequent month she adds \(\,€50\). The account earns no interest. We model the balance at the end of each month as an arithmetic progression.
Let \(a_n\) denote the balance, in euros, at the end of month \(n\). At the end of month 1 the balance is \(200 + 50 = 250\), so \(a_1 = 250\). Each subsequent month the balance grows by \(50\), so the common difference is \(d = 50\). The general term is
\[a_n = 250 + (n - 1) \cdot 50 = 50\,n + 200.\]
Balance after one year. Setting \(n = 12\), \(a_{12} = 50 \cdot 12 + 200 = 800\). At the end of December the account holds \(\,€800\).
When does the balance reach €1000? We solve \(a_n \geq 1000\), that is \(50\,n + 200 \geq 1000\), giving \(n \geq 16\). The threshold is first met at the end of month 16 — four months into the second year.
A training plan
A runner plans a training programme in which she covers \(3\) kilometres on the first day and adds \(200\) metres to her run each subsequent day. We ask two questions: how far does she run on the thirtieth day, and how far has she run in total over those thirty days?
Working in kilometres, let \(a_n\) be the distance run on day \(n\). Then \(a_1 = 3\) and the common difference is \(d = 0.2\). By the general term,
\[a_{30} = 3 + 29 \cdot 0.2 = 3 + 5.8 = 8.8 \text{ km}.\]
For the total, the partial sum formula gives
\[S_{30} = \frac{30 \cdot (a_1 + a_{30})}{2} = \frac{30 \cdot (3 + 8.8)}{2} = 15 \cdot 11.8 = 177 \text{ km}.\]
The single-day distance grows linearly, but the cumulative distance — being the partial sum — grows quadratically. Doubling the number of training days more than doubles the total distance run.
Straight-line depreciation
A laptop is purchased new for \(\,€1200\). The accountant assumes it loses \(\,€100\) of book value (the value at which the asset is carried on the company's accounts) each year — a method called straight-line depreciation. Let \(V_n\) denote the book value at the end of year \(n\), and take \(V_0 = 1200\) as the purchase value. Each year subtracts \(100\), so
\[V_n = 1200 - 100\,n.\]
This is an arithmetic progression with \(d = -100\), indexed from \(n = 0\). The book value reaches zero when \(1200 - 100\,n = 0\), that is, after \(n = 12\) years. Beyond that point the linear model would predict negative values, which has no accounting meaning; in practice the book value is held at zero (or at a small salvage value) once the depreciation schedule is exhausted.
This example illustrates a general lesson: the mathematics of the arithmetic progression is happy to continue forever, but the real-world quantity being modelled need not be. Part of using a model responsibly is knowing where it ceases to apply.
Connection to Computer Science
The same modelling moves appear in the analysis of programs. A scheduler that allocates a fixed time slice to each iteration of a loop accumulates time as an arithmetic progression: total time after \(n\) iterations is \(n\) times the slice, plus any fixed overhead. A loop that processes a fixed number of bytes per iteration produces a balance that grows linearly with iteration count — exactly Maria's account in another guise.
The training-plan example, where the partial sum grows quadratically while the per-step term grows linearly, is the smallest instance of a phenomenon that pervades algorithm analysis. A doubly nested loop in which the inner loop runs \(1, 2, 3, \ldots, n\) times performs \(\dfrac{n(n+1)}{2}\) operations in total — quadratic running time built from linearly growing inner work. The sum formula tells you immediately why such algorithms are slow on large inputs.
When linear is the wrong model
Imagine Maria's savings account instead earns interest at a fixed annual rate, compounded monthly. Now the addition to her balance each month is not a fixed \(\,€50\) but a small percentage of whatever is already in the account. The balance still grows, but by a larger amount each month, because the base on which the interest is computed has itself grown. The model is no longer an arithmetic progression.
Medieval and Renaissance commercial mathematics drew this distinction carefully. Simple interest — the linear case, modelled by an arithmetic progression — was the calculation of choice for short-term loans and modest sums. Compound interest, in which the interest itself earns interest, is what allows capital to grow without bound; the systematic treatment of both regimes appears already in Luca Pacioli's Summa de Arithmetica of 1494, the source from which much of modern double-entry bookkeeping descends. The mathematics of compound interest is the mathematics of geometric progressions, and we develop it in the next unit.
Exercises
Exercise 1
Ana starts a savings plan on the first of February with an initial deposit of €300 and adds €80 at the end of each subsequent month. Let \(a_n\) be the balance in euros at the end of month \(n\), so \(a_1 = 380\). The account earns no interest.
Exercise 2
A cyclist's training plan: \(20\) kilometres on day 1, with \(2\) kilometres added each subsequent day. Distances are measured in kilometres.
Exercise 3
A machine is purchased for €18 000. The accounting department uses straight-line depreciation, writing down €1 500 of book value at the end of each year. Let \(V_n\) be the book value at the end of year \(n\), with \(V_0 = 18000\).
Exercise 4
Decide whether each situation is naturally modelled by an arithmetic progression.
Summary
Arithmetic progressions arise naturally whenever a quantity changes by the same amount over equal intervals: savings with fixed contributions, training distances that grow by a fixed step, equipment depreciating by a fixed amount each year.
To build the model, identify the quantity \(a_n\), the step, the initial value \(a_1\), the common difference \(d\), and the question being asked. The general-term formula handles single values; the partial-sum formula handles totals; an inequality handles threshold questions.
A linear model is right precisely when the change per step is constant. If the change is instead proportional to the current value — interest on capital, doubling populations — the right tool is a geometric progression.
A model is a tool, not a fact: when a depreciating book value would become negative, the model has reached the boundary of its meaning, and the question shifts from mathematics to accounting policy.