Number & Algebra — IB Math AI

1. Number Systems and Scientific Notation

AI students work extensively with large and small numbers in real-world contexts. Scientific notation and significant figures are foundational skills.

Scientific Notation and Significant Figures

Scientific notation: $a \times 10^n$ where $1 \le a < 10$ and $n \in \mathbb{Z}$

Rounding rules: Identify the last significant digit; if the next digit is 5 or more, round up.

Percentage error: $\dfrac{|v_A - v_E|}{|v_E|} \times 100\%$ where $v_A$ is approximate and $v_E$ is exact.

2. Sequences and Series in Financial Contexts

Arithmetic and Geometric Sequences

Arithmetic: $u_n = u_1 + (n-1)d$, $S_n = \dfrac{n}{2}(2u_1 + (n-1)d) = \dfrac{n}{2}(u_1+u_n)$

Geometric: $u_n = u_1 r^{n-1}$, $S_n = \dfrac{u_1(r^n-1)}{r-1}$ for $r \ne 1$, $S_\infty = \dfrac{u_1}{1-r}$ for $|r|<1$

3. Financial Mathematics

This is a major focus of AI Topic 1. Students use GDC financial applications to solve problems involving savings, loans, and investments.

Compound Interest and Present/Future Value

Compound interest: $FV = PV\!\left(1+\dfrac{r}{100n}\right)^{nt}$

where $PV$ = present value, $r$ = annual interest rate (%), $n$ = compounding periods per year, $t$ = time in years.

Annual effective rate: $r_{\text{eff}} = \left(1+\dfrac{r}{100n}\right)^n - 1$

Worked Example 1

Compound Interest: Investment Growth

Anita invests \$5000 at 4.5% per year, compounded monthly. How much does she have after 8 years?

Identify values: $PV = 5000$, $r = 4.5$, $n = 12$ (monthly), $t = 8$

Apply formula: $FV = 5000\!\left(1+\dfrac{4.5}{100 \times 12}\right)^{12 \times 8} = 5000(1.00375)^{96}$

Calculate: $(1.00375)^{96} \approx 1.4326$, so $FV \approx 5000 \times 1.4326 \approx \$7163.10$

Loans and Amortization

For a loan of amount $L$ at monthly interest rate $i = r/(100 \times 12)$ paid over $N$ monthly payments, the regular payment $P$ is:

$$P = \frac{Li}{1-(1+i)^{-N}}$$

In the IB AI course, the GDC TVM (Time Value of Money) solver is used for all financial calculations. You must know which values to enter into $N$, $I\%$, $PV$, $PMT$, and $FV$.

Worked Example 2

Loan Repayment: Monthly Installments

Ben takes a \$15,000 car loan at 6% per year, compounded monthly, to be repaid over 5 years. Find the monthly payment and total interest paid.

Parameters: $L=15000$, $i = 6/(100\times12) = 0.005$, $N = 60$ months

Monthly payment: $P = \dfrac{15000 \times 0.005}{1-(1.005)^{-60}} = \dfrac{75}{1-0.7414} = \dfrac{75}{0.2586} \approx \$289.99$

Total paid: $60 \times 289.99 = \$17,399.40$. Total interest $= 17399.40 - 15000 = \$2399.40$.

Worked Example 3

Annuity: Regular Savings Plan

Clara saves \$200 per month in an account earning 3% per year, compounded monthly. How much does she have after 10 years?

Geometric series approach: Each payment earns compound interest. Monthly rate $i = 0.0025$, $N = 120$ payments.

Future value of annuity: $FV = PMT \times \dfrac{(1+i)^N - 1}{i} = 200 \times \dfrac{(1.0025)^{120}-1}{0.0025}$

Calculate: $(1.0025)^{120} \approx 1.3494$, so $FV = 200 \times \dfrac{0.3494}{0.0025} = 200 \times 139.76 \approx \$27,952$

GDC Note (AI Students): Use the TVM solver on your GDC for all financial calculations. Set $PV$ as negative (outflow) and $FV$ as positive (inflow) or vice versa depending on the context. The GDC handles the sign convention automatically in most models.

Practice Problems

Q1. An investment of \$3000 grows to \$4500 over 6 years with annual compounding. Find the annual interest rate.

Show Solution

$4500 = 3000(1+r/100)^6 \Rightarrow (1+r/100)^6 = 1.5$

$1+r/100 = 1.5^{1/6} = 1.0699$ so $r \approx 6.99\%$

Q2. A mortgage of \$250,000 is taken at 5% annual interest, compounded monthly, for 25 years. Find the monthly payment.

Show Solution

$i = 0.05/12 \approx 0.004167$, $N = 300$.

$P = \dfrac{250000 \times 0.004167}{1-(1.004167)^{-300}} = \dfrac{1041.7}{1-0.2873} \approx \dfrac{1041.7}{0.7127} \approx \$1461.88$

Q3. A sequence of annual salaries starts at \$45,000 and increases by 3% each year. What is the total earned over 10 years?

Show Solution

Geometric series: $u_1 = 45000$, $r = 1.03$, $n = 10$.

$S_{10} = \dfrac{45000(1.03^{10}-1)}{0.03} = \dfrac{45000(1.3439-1)}{0.03} = \dfrac{45000 \times 0.3439}{0.03} \approx \$515,850$

4. Exam Tips

Always state whether interest is compounded annually, monthly, or daily — this changes $n$ in the formula.
When using GDC TVM, set $P/Y$ and $C/Y$ correctly (payments per year and compounding periods per year).
Percentage error questions require careful identification of which value is "exact" and which is "approximate."
For loans: the outstanding balance at any point can be found using TVM with the remaining number of payments.