4️⃣ Introduction to Financial-Mathematics Tools
Study numerical sequences — general definition, arithmetic sequences, and geometric sequences — in order to understand and model phenomena that evolve regularly over time, such as savings, loans, and price indices.
- 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.
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.
- 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
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 \(r = 100\).
\[ u_n = u_0 + n\,r. \] 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, \(u_{12} = 500 + 100 \times 12 = 1700\) €.
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
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 \(q = 2\).
\[ u_n = u_0 \times q^{\,n}. \] For \(u_0 > 0\): the sequence is increasing if \(q > 1\), decreasing if \(0 < q < 1\), and constant if \(q = 1\).
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 \(q \neq 1\),
\[ S = u_0 \,\frac{1 - q^{\,n+1}}{1 - q} = (\text{first term}) \times \frac{1 - q^{\,\text{number of terms}}}{1 - q}. \]
Synthesis Exercise — two salary-growth options
An employee is hired at an annual salary of \(28{,}000\) €. The employer offers two options for how the salary will evolve:
- Option A — fixed increase: each year the salary rises by \(1{,}500\) €.
- Option B — percentage increase: each year the salary rises by \(5\%\) relative to the previous year.
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.
Practice
- An arithmetic sequence has \(u_0 = 200\) and \(r = 15\). Give \(u_{10}\) and the sum \(u_0 + \dots + u_{10}\). \(u_{10} = 200 + 150 = 350\); \(S = 11 \times \tfrac{200 + 350}{2} = 3025\).
- A geometric sequence has \(u_0 = 1000\) and \(q = 1.02\). Give \(u_5\) and the sum of the first \(6\) terms. \(u_5 = 1000 \times 1.02^{5} \approx 1104.08\); \(S = 1000 \times \tfrac{1 - 1.02^{6}}{1 - 1.02} \approx 6308.12\).
TODO: interactive sequence explorer (arithmetic vs geometric growth) deferred — see plan Complexity Tracking.