Introduction to Financial-Mathematics Tools

NoteLearning objectives

• Understand what a numerical sequence is and use index notation.
• Recognise, model, and study arithmetic sequences.
• Recognise, model, and study geometric sequences.
• Compute the sum of the first terms of each type of sequence.

📊 Lecture slides: View the slides for this module

1 Diagnostic Activity

Answer each question from memory before reading further.

1. What is the next term in the sequence 3, 7, 11, 15, …?
2. What is the next term in the sequence 2, 6, 18, 54, …?
3. Is the sequence 100, 50, 25, 12.5 arithmetic or geometric?

NoteCheck your answers

1. 19 — each term increases by 4.
2. 162 — each term is multiplied by 3.
3. Geometric, with common ratio \(q = 0.5\).

Use your results to identify which sequence type needs more attention.

2 Opening Problem

Two colleagues are each offered a salary raise starting from €28,000 per year. Option A: a fixed raise of €1,500 per year. Option B: a 5% raise each year relative to the previous year’s salary.

After 10 years, which colleague earns more? Write down your prediction before continuing.

3 Core Concepts

Numerical Sequences

A numerical sequence is an ordered list of numbers indexed by the whole numbers. The term of index \(n\) is written \(u_n\); \(u_0\) (or \(u_1\)) is the first term. Sequences are used to model phenomena that evolve regularly — that is, over a constant period we add or multiply by the same value.

TipWhere sequences appear in economics and management

• Savings or loan repayment: a capital that grows or shrinks month by month.
• Accounting: computing annual depreciation.
• Management: forecasting when equipment must be replaced.
• Macroeconomics: tracking a price index over time.

A sequence can be defined in two ways:

• recursively, giving \(u_0\) and a rule linking \(u_{n+1}\) to \(u_n\);
• explicitly, giving \(u_n\) directly as a function of \(n\).

Arithmetic Sequences

NoteDefinition

A sequence is arithmetic if each term is obtained by adding a constant number \(r\), called the common difference, to the previous term: \[ u_{n+1} = u_n + r. \]

Introductory example. A person saves \(100\) € per month (without interest), added to an initial capital of \(500\) €. Each month we add the constant amount \(100\) €, so the saved amount forms an arithmetic sequence with first term \(u_0 = 500\) and common difference \(r = 100\).

TipExplicit form and sense of variation

\[ u_n = u_0 + n\,r. \] The sequence is increasing if \(r > 0\), decreasing if \(r < 0\), and constant if \(r = 0\).

For the savings example, \(u_n = 500 + 100\,n\). After \(12\) months, \(u_{12} = 500 + 100 \times 12 = 1700\) €.

CautionDerivation — sum of the first terms of an arithmetic sequence (click)

Let \(S = u_0 + u_1 + \dots + u_n\) (there are \(n + 1\) terms). Write the sum forwards and backwards and add termwise (Gauss’s trick):

\[ \begin{aligned} S &= u_0 + u_1 + \dots + u_n, \\ S &= u_n + u_{n-1} + \dots + u_0. \end{aligned} \]

Each vertical pair sums to \(u_0 + u_n\) (because moving up on one line by \(r\) is compensated by moving down on the other), and there are \(n + 1\) such pairs, so \(2S = (n + 1)(u_0 + u_n)\). Hence

\[ S = \frac{(n + 1)(u_0 + u_n)}{2} = (\text{number of terms}) \times \frac{\text{first} + \text{last}}{2}. \]

Geometric Sequences

NoteDefinition

A sequence is geometric if each term is obtained by multiplying the previous term by a constant number \(q\), called the common ratio: \[ u_{n+1} = u_n \times q. \]

Introductory example. Folding a sheet of paper doubles its thickness at every fold: at each fold we multiply by the constant \(2\). The thickness forms a geometric sequence with common ratio \(q = 2\).

TipExplicit form and sense of variation

\[ u_n = u_0 \times q^{\,n}. \] For \(u_0 > 0\): the sequence is increasing if \(q > 1\), decreasing if \(0 < q < 1\), and constant if \(q = 1\).

CautionDerivation — sum of the first terms of a geometric sequence (click)

Let \(S = u_0 + u_0 q + u_0 q^{2} + \dots + u_0 q^{n}\). Multiply by \(q\): \(qS = u_0 q + u_0 q^{2} + \dots + u_0 q^{n+1}\). Subtracting,

\[ S - qS = u_0 - u_0 q^{n+1} \;\Longrightarrow\; S(1 - q) = u_0\,(1 - q^{\,n+1}). \]

Therefore, for \(q \neq 1\),

\[ S = u_0 \,\frac{1 - q^{\,n+1}}{1 - q} = (\text{first term}) \times \frac{1 - q^{\,\text{number of terms}}}{1 - q}. \]

CautionExercises (click for solutions)

1. An arithmetic sequence has \(u_0 = 200\) and \(r = 15\). Give \(u_{10}\) and the sum \(u_0 + \dots + u_{10}\). \(u_{10} = 200 + 150 = 350\); \(S = 11 \times \tfrac{200 + 350}{2} = 3025\).
2. A geometric sequence has \(u_0 = 1000\) and \(q = 1.02\). Give \(u_5\) and the sum of the first \(6\) terms. \(u_5 = 1000 \times 1.02^{5} \approx 1104.08\); \(S = 1000 \times \tfrac{1 - 1.02^{6}}{1 - 1.02} \approx 6308.12\).

4 Worked Example

Worked example — two salary-growth options

An employee starts at €28,000/year. Compare a fixed raise of €1,500/year (Option A) vs. a 5% annual raise (Option B).

1. Nature of each sequence. Option A adds a constant amount → arithmetic sequence (\(u_0 = 28{,}000\), \(r = 1{,}500\)). Option B multiplies by a constant factor → geometric sequence (\(v_0 = 28{,}000\), \(q = 1.05\)).

2. Salary after 6 years (index \(6\), the 7th year):

• Option A: \(u_6 = 28{,}000 + 1{,}500 \times 6 = 37{,}000\) €.
• Option B: \(v_6 = 28{,}000 \times 1.05^{6} \approx 37{,}527\) €.

3. Cumulative salary over the first 6 years (indices \(0\) to \(5\), i.e. \(6\) terms):

• Option A: \(S_A = 6 \times \dfrac{u_0 + u_5}{2} = 6 \times \dfrac{28{,}000 + 35{,}500}{2} = 190{,}500\) €.
• Option B: \(S_B = 28{,}000 \times \dfrac{1 - 1.05^{6}}{1 - 1.05} \approx 190{,}570\) €.

4–6. Comment. Option A gives a higher annual salary in the early years, but Option B grows faster and overtakes Option A after a few years because geometric (percentage) growth eventually dominates arithmetic (fixed) growth. The crossover is read from the table of yearly values.

5 Guided Practice

A company purchases equipment worth €50,000. The equipment depreciates by €4,000 per year (straight-line). A rival company’s equipment depreciates by 12% of its remaining value each year, starting from the same price.

1. Write the value function \(u_n\) for each company after \(n\) years.
2. Compute the value for each company after 5 years.
3. After how many years does Company B’s equipment first fall below €20,000?

TipHint

Company A: arithmetic with \(u_0 = 50{,}000\) and \(r = -4{,}000\). Company B: geometric with \(u_0 = 50{,}000\) and \(q = 0.88\).

CautionShow the solution

• Company A: \(u_n = 50{,}000 - 4{,}000n\). At \(n = 5\): \(u_5 = 30{,}000\) €.
• Company B: \(u_n = 50{,}000 \times 0.88^n\). At \(n = 5\): \(u_5 = 50{,}000 \times 0.88^5 \approx 27{,}442\) €.
• For Company B below €20,000: \(50{,}000 \times 0.88^n < 20{,}000 \Rightarrow 0.88^n < 0.4\). At \(n = 7\): \(0.88^7 \approx 0.409\); at \(n = 8\): \(0.88^8 \approx 0.360 < 0.4\). So after 8 years.

6 Interactive Exploration

Use the sliders below to explore how arithmetic and geometric sequences compare as you vary the initial term, common difference, and common ratio.

viewof u0 = Inputs.range([0, 100], {
  value: 28,
  step: 1,
  label: "Initial term u₀"
})

viewof d = Inputs.range([-10, 10], {
  value: 1.5,
  step: 0.5,
  label: "Common difference d (arithmetic)"
})

viewof r = Inputs.range([0.5, 2.0], {
  value: 1.05,
  step: 0.01,
  label: "Common ratio r (geometric)"
})

viewof nTerms = Inputs.range([2, 20], {
  value: 10,
  step: 1,
  label: "Number of terms n"
})

indices = Array.from({length: nTerms}, (_, i) => i)
arithmeticSeq = indices.map(n => ({n, value: u0 + n * d, type: "Arithmetic"}))
geometricSeq = indices.map(n => ({n, value: u0 * Math.pow(r, n), type: "Geometric"}))
allData = [...arithmeticSeq, ...geometricSeq]

seqChart = Plot.plot({
  marks: [
    Plot.line(allData, {x: "n", y: "value", stroke: "type"}),
    Plot.dot(allData, {x: "n", y: "value", fill: "type"})
  ],
  x: {label: "Term index n"},
  y: {label: "Value"},
  color: {legend: true}
})

html`
<div style="margin-bottom:1rem">${seqChart}</div>
<div class="metric-grid">
  <div class="metric-card">
    <span class="metric-label">Arithmetic u₍ₙ₎ (last term)</span>
    <strong>${(u0 + (nTerms-1)*d).toFixed(2)}</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Geometric u₍ₙ₎ (last term)</span>
    <strong>${(u0 * Math.pow(r, nTerms-1)).toFixed(2)}</strong>
  </div>
</div>
`

Investigation

1. Set \(u_0 = 28\), \(d = 1.5\), \(r = 1.05\), \(n = 10\). Which sequence is larger at the last term?
2. Increase \(n\) to 20. At what point does the geometric sequence overtake the arithmetic one?
3. Set \(r = 0.88\) and \(d = -4\). Which model represents faster depreciation?

7 Applied Problem

A price index tracks the cost of a basket of goods. In 2020 the index stands at 100. Over the following 5 years, prices increase by a fixed €2 points per year (Scenario A) or by 2% per year (Scenario B).

1. Write the index value \(I_n\) for each scenario after \(n\) years (\(n = 0\) is 2020).
2. Compute the index value for each scenario after 5 years.
3. Which scenario leads to a higher cumulative price increase over 10 years?
4. Interpret: what does an index above 100 mean for consumers?

8 Common Errors

WarningCommon error — mixing up common difference and common ratio

A student is given the sequence 3, 6, 12, 24 and writes \(u_n = 3 + n \times 3\).

This is incorrect — the sequence is geometric (each term is multiplied by 2), not arithmetic (which adds a constant). The correct formula is \(u_n = 3 \times 2^n\).

Always check whether consecutive terms have a constant difference (arithmetic) or a constant ratio (geometric) before applying a formula.

9 Check Your Understanding

1. A sequence has \(u_0 = 5\) and \(u_{n+1} = u_n + 3\). What is \(u_6\)? What is the sum \(u_0 + u_1 + \dots + u_6\)?
2. Why does a geometric sequence with ratio \(r > 1\) grow faster than any arithmetic sequence with the same first term, for large enough \(n\)?
3. A savings account earns 4% interest per year (compound). Modelled as a geometric sequence, what is the common ratio?
4. A student claims the sum of the first 5 terms of the arithmetic sequence \(2, 5, 8, 11, 14\) is \(5 \times 8 = 40\) (using the average). Is this correct?
5. Two sequences both start at \(u_0 = 100\). Sequence A has \(d = 10\); Sequence B has \(r = 1.08\). After how many terms does Sequence B first exceed Sequence A?

10 Summary

1. An arithmetic sequence grows by a constant amount each period (\(u_n = u_0 + nr\)); a geometric sequence grows by a constant factor (\(u_n = u_0 r^n\)) — recognising which model applies is the first step in any analysis.
2. For the same starting value and similar short-term growth, geometric growth always eventually dominates arithmetic growth as \(n\) increases — this is why compound interest outperforms simple interest over long horizons.
3. Sum formulas — arithmetic: \(S_n = (n+1)(u_0 + u_n)/2\); geometric: \(S_n = u_0(1 - r^{n+1})/(1-r)\) — let us aggregate total accumulations over time without summing term by term.

11 Independent Work

• Complete sequence exercises 1–10 in the exercise bank.
• Work through the salary comparison case study.
• Record errors in your error log.
• Review glossary entries for: arithmetic sequence, geometric sequence, common difference, common ratio, sum formula.

12 Continue

Continue to Module 5 →