IB SL

Number and Algebra

IB Math AI SL — Number and Algebra

Complete Study Guide

Topics Covered

Scientific Notation and Estimation — rounding, significant figures, percentage error
Arithmetic Sequences and Series — modelling linear growth, sigma notation
Geometric Sequences and Series — modelling exponential growth and decay
Financial Mathematics — compound interest, annuities, amortization
Logarithms in Context — solving exponential equations, half-life, doubling time
Practice Questions and Exam Alerts

Topic 1 of the IB Math AI SL syllabus — Paper 1 and Paper 2

Calculator-active course: Unlike Math AA, all AI papers allow GDC use. You still need to show working for “show that” questions, but for most problems your GDC is your best friend. Know how to use the finance solver (TVM), table of values, and equation solver efficiently.

Core formulas at a glance

Formula	Expression
Percentage error	$\left\lvert\dfrac{v_A - v_E}{v_E}\right\rvert \times 100\%$
Arithmetic $n$ th term	$u_n = u_1 + (n-1)d$
Arithmetic sum	$S_n = \dfrac{n}{2}(2u_1 + (n-1)d)$
Geometric $n$ th term	$u_n = u_1 \cdot r^{n-1}$
Geometric sum	$S_n = \dfrac{u_1(1 - r^n)}{1 - r},\ r \neq 1$
Compound interest	$FV = PV \times \left(1 + \dfrac{r}{100k}\right)^{kn}$

Section 1: Scientific Notation, Rounding, and Estimation

1.1 Scientific Notation

Any number can be written in the form $a \times 10^k$ where $1 \le a < 10$ and $k$ is an integer.

Real-world context: Scientists measuring cell sizes ( $3.2 \times 10^{-6}$ m) or national GDPs ( $1.8 \times 10^{12}$ USD) use scientific notation to keep numbers manageable. On Paper 2, data is frequently given in this form.

Scientific notation in context

The population of bacteria in a culture doubles every 3 hours. Starting from 500 bacteria, express the population after 24 hours in scientific notation.

After 24 hours there are $\frac{24}{3} = 8$ doubling periods.

$P = 500 \times 2^8 = 500 \times 256 = 128\,000 = 1.28 \times 10^5$

1.2 Significant Figures and Rounding

The IB expects you to give final answers to 3 significant figures unless told otherwise. Intermediate calculations should carry at least one extra significant figure to avoid rounding error.

Percentage error measures how far an approximation $v_A$ is from an exact value $v_E$ :

$\text{Percentage error} = \left\lvert \frac{v_A - v_E}{v_E} \right\rvert \times 100\%$

Common mistake: Students sometimes use the approximation in the denominator. The IB formula uses the exact value in the denominator. This is given in the formula booklet.

Percentage error

A student estimates the height of a building as 14.5 m. The actual height is 15.2 m. Find the percentage error.

$\text{Percentage error} = \left\lvert \frac{14.5 - 15.2}{15.2} \right\rvert \times 100\% = \frac{0.7}{15.2} \times 100\% = 4.61\%$

Section 2: Arithmetic Sequences and Series

An arithmetic sequence has a constant difference $d$ between consecutive terms. This models linear growth: salary increases by a fixed amount each year, a taxi charges a fixed rate per km.

2.1 The $n$ th Term

$u_n = u_1 + (n-1)d$

where $u_1$ is the first term and $d$ is the common difference.

2.2 The Sum of $n$ Terms

$S_n = \frac{n}{2}(2u_1 + (n-1)d) = \frac{n}{2}(u_1 + u_n)$

Arithmetic sequence in a real-world scenario

A gym membership costs $40 in the first month, and increases by$ 5 each month. Find the cost in month 12 and the total cost over the first 12 months.

Here $u_1 = 40$ , $d = 5$ .

Month 12: $u_{12} = 40 + 11 \times 5 = 40 + 55 = 95$ dollars.

Total cost: $S_{12} = \dfrac{12}{2}(40 + 95) = 6 \times 135 = 810$ dollars.

2.3 Applications

Arithmetic sequences model situations where a quantity changes by the same amount each period:

Saving a fixed amount each month
Depreciation by a fixed dollar value
Seating in a stadium (each row has $d$ more seats than the previous row)

GDC shortcut: On TI-84, enter the sequence formula in $Y_1$ and use TABLE to see all terms at once. On Casio, use the sequence mode (RECUR). This is much faster than calculating each term by hand.

Section 3: Geometric Sequences and Series

A geometric sequence has a constant ratio $r$ between consecutive terms. This models exponential growth and decay: population growth, radioactive decay, compound interest.

3.1 The $n$ th Term

$u_n = u_1 \cdot r^{n-1}$

3.2 The Sum of $n$ Terms

$S_n = \frac{u_1(1 - r^n)}{1 - r}, \quad r \neq 1$

3.3 Infinite Geometric Series

When $|r| < 1$ , the sum to infinity converges:

$S_\infty = \frac{u_1}{1 - r}$

Geometric sequence — population decline

A town has 12,000 residents. Each year the population decreases by 3%. Find the population after 10 years and the total number of “person-years” lived over those 10 years.

Here $u_1 = 12000$ , $r = 0.97$ .

After 10 years: $u_{11} = 12000 \times 0.97^{10} = 12000 \times 0.7374\ldots = 8849$ (3 s.f.)

Total person-years: $S_{10} = \dfrac{12000(1 - 0.97^{10})}{1 - 0.97} = \dfrac{12000 \times 0.2626}{0.03} = 105\,000$ (3 s.f.)

Be careful with $n$ : If the population starts at 12,000, this is $u_1$ . The population after 1 year is $u_2 = u_1 \cdot r$ . After 10 years is $u_{11}$ . Many students use $n = 10$ when they should use $n = 11$ .

Section 4: Financial Mathematics

Financial maths is a major focus of Math AI. You must understand compound interest and be proficient with your GDC’s finance solver (TVM solver).

4.1 Compound Interest

$FV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}$

where:

$FV$ = future value
$PV$ = present value (initial investment)
$r$ = annual interest rate (as a percentage)
$k$ = number of compounding periods per year
$n$ = number of years

Compounding	$k$
Annually	1
Quarterly	4
Monthly	12
Daily	365

Compound interest — comparing options

Anna invests $5000 at 4.2% per annum compounded monthly. Ben invests$ 5000 at 4.3% per annum compounded annually. Who has more after 5 years?

Anna: $FV = 5000 \times \left(1 + \dfrac{4.2}{1200}\right)^{60} = 5000 \times 1.0035^{60} = 5000 \times 1.2330 = 6165$ (nearest dollar)

Ben: $FV = 5000 \times \left(1 + \dfrac{4.3}{100}\right)^{5} = 5000 \times 1.043^{5} = 5000 \times 1.2349 = 6175$ (nearest dollar)

Ben has slightly more. The higher nominal rate outweighs monthly compounding.

4.2 Using the GDC Finance Solver (TVM)

Your GDC has a TVM (Time Value of Money) solver. The key variables are:

Variable	Meaning
$N$	Total number of payment periods
$I\%$	Annual interest rate
$PV$	Present value (negative if you pay it)
$PMT$	Payment per period
$FV$	Future value
$P/Y$	Payments per year
$C/Y$	Compounding periods per year

Sign convention: Money you pay out is negative, money you receive is positive. For a loan, $PV$ is positive (you receive the money) and $PMT$ is negative (you pay it back).

4.3 Annuities and Amortization

An annuity is a series of equal payments made at regular intervals. An amortization schedule shows how each payment is split between interest and principal.

Loan repayment

You borrow 20,000 USD at 6% annual interest compounded monthly. You repay in equal monthly instalments over 5 years. Find the monthly payment.

Using the TVM solver: $N = 60$ , $I\% = 6$ , $PV = 20000$ , $FV = 0$ , $P/Y = 12$ , $C/Y = 12$ .

Solve for $PMT$ : $PMT = -386.66$ (the negative sign confirms you are paying out).

Monthly payment is 386.66 USD.

Total paid: $386.66 \times 60 = 23\,200$ . Total interest: $23\,200 - 20\,000 = 3200$ .

Section 5: Logarithms in Context

Logarithms in Math AI focus on solving real-world exponential equations rather than algebraic manipulation.

5.1 The Logarithm as an Inverse

If $a^x = b$ then $x = \log_a b$ .

The most common bases are 10 ( $\log$ ) and $e$ ( $\ln$ ).

5.2 Solving Exponential Equations

To find when a quantity reaches a target value, take logarithms of both sides:

$u_1 \cdot r^n = T \implies n = \frac{\ln T - \ln u_1}{\ln r}$

Doubling time

A population of 8000 grows at 2.5% per year. How long until it reaches 16,000?

$8000 \times 1.025^n = 16000$ $1.025^n = 2$ $n = \frac{\ln 2}{\ln 1.025} = \frac{0.6931}{0.02469} = 28.1 \text{ years}$

5.3 Half-Life

The time for a quantity to halve. If $r$ is the decay factor per unit time:

$t_{1/2} = \frac{\ln 0.5}{\ln r}$

Radioactive decay

A sample of 500 g of a radioactive substance decays by 8% per year. Find the half-life and the mass after 15 years.

Decay factor $r = 0.92$ .

Half-life: $t_{1/2} = \dfrac{\ln 0.5}{\ln 0.92} = \dfrac{-0.6931}{-0.08338} = 8.31$ years.

After 15 years: $500 \times 0.92^{15} = 500 \times 0.2863 = 143$ g (3 s.f.)

Section 6: Practice Questions

Paper 1 Style (Short Answer)

Q1. The 5th term of an arithmetic sequence is 22 and the 12th term is 57. Find

u_1

and

d

From $u_5 = u_1 + 4d = 22$ and $u_{12} = u_1 + 11d = 57$ :

Subtracting: $7d = 35$ , so $d = 5$ .

$u_1 = 22 - 4(5) = 2$ .

Q2. Express 0.000427 in scientific notation and find the percentage error when it is rounded to

4.3 \times 10^{-4}

$0.000427 = 4.27 \times 10^{-4}$ .

Percentage error: $\left\lvert\dfrac{4.3 \times 10^{-4} - 4.27 \times 10^{-4}}{4.27 \times 10^{-4}}\right\rvert \times 100\% = \dfrac{0.03}{4.27} \times 100\% = 0.702\%$ .

Q3. A car purchased for 25,000 depreciates by 15% each year. Find its value after 6 years.

$V = 25000 \times 0.85^6 = 25000 \times 0.3771 = 9428$ (nearest dollar).

Paper 2 Style (Extended Response)

Q4. Sophia invests 8000 at 3.8% per annum compounded quarterly. (a) Find the value after 10 years. (b) Find when the investment first exceeds 12,000. (c) If instead she adds 100 per quarter, use your GDC to find the value after 10 years.

(a) $FV = 8000 \times \left(1 + \dfrac{3.8}{400}\right)^{40} = 8000 \times 1.0095^{40} = 8000 \times 1.4596 = 11\,677$ (nearest dollar).

(b) $8000 \times 1.0095^{4n} = 12000$

$1.0095^{4n} = 1.5$

$4n = \dfrac{\ln 1.5}{\ln 1.0095} = \dfrac{0.4055}{0.009455} = 42.89$

$n = 10.7$ years, so in the 11th year.

(c) Using GDC TVM: $N = 40$ , $I\% = 3.8$ , $PV = -8000$ , $PMT = -100$ , $P/Y = 4$ , $C/Y = 4$ . Solve for $FV$ : $FV = 16\,518$ (nearest dollar).

Q5. A bouncing ball is dropped from 2 metres. After each bounce it reaches 72% of its previous height. (a) Find the height after the 5th bounce. (b) Find the total vertical distance the ball travels before it stops.

(a) After the 5th bounce: $h_5 = 2 \times 0.72^5 = 2 \times 0.1935 = 0.387$ m.

(b) The ball falls 2 m, then bounces up and down. Total distance:

$D = 2 + 2 \times S_\infty$ where $S_\infty = \dfrac{2 \times 0.72}{1 - 0.72} = \dfrac{1.44}{0.28} = 5.143$ m.

$D = 2 + 2(5.143) = 12.3$ m (3 s.f.).

Exam strategy for financial maths: Always write down the TVM values you enter, even if you solve on the GDC. The IB expects to see $N = ..., I\% = ..., PV = ..., PMT = ..., FV = ..., P/Y = ..., C/Y = ...$ before the answer. Without this, you lose method marks.

May 2026 Prediction Questions

These are NOT official IB questions. These are trend-based practice questions written to reflect the topic areas and question styles most likely to appear on the May 2026 IB Math AI SL Paper 2. Based on recent exam patterns (2022–2025), expect heavy weighting on: compound interest, annuities (TVM solver), geometric series in real-world contexts, and logarithms in exponential growth/decay scenarios.

Question 1 — Financial Mathematics (TVM Solver) [~8 marks]

Priya takes out a car loan of $18,500 at an annual interest rate of 5.4%, compounded monthly. She makes equal monthly repayments over 4 years.

(a) Find the monthly repayment amount.

(b) Find the total amount Priya pays over the 4 years.

(d) After 2 years of payments, Priya wants to pay off the remaining balance in full. Using your GDC, find the outstanding balance at that point.

Show Solution

(a) Set up TVM solver:

$N = 48$ , $I\% = 5.4$ , $PV = 18500$ , $FV = 0$ , $P/Y = 12$ , $C/Y = 12$ .

Solve for $PMT$ : $PMT = -429.40$

Monthly repayment is $429.40.

(b) Total paid $= 429.40 \times 48 = 20\,611$ (nearest dollar, USD).

(c) Total interest $= 20611 - 18500 = 2111$ (USD).

(d) After 2 years, 24 payments have been made. Use TVM:

$N = 24$ , $I\% = 5.4$ , $PV = 18500$ , $PMT = -429.40$ , $P/Y = 12$ , $C/Y = 12$ .

Solve for $FV$ : $FV = -9748$ (magnitude is the outstanding balance).

Outstanding balance after 2 years: $9748.

Question 2 — Geometric Series in Context [~7 marks]

A biologist studying a bacterial culture observes that the population triples every 90 minutes. The initial population is 400 bacteria.

(a) Write a model for the population $P$ after $n$ intervals of 90 minutes.

(b) Find the population after 6 hours.

(d) The culture vessel has a capacity of 5,000,000 bacteria. Find the total number of bacteria across the first 8 intervals and comment on whether the vessel is sufficient.

Show Solution

(a) This is a geometric sequence with $u_1 = 400$ and $r = 3$ .

$P(n) = 400 \times 3^{n-1}$

where $n$ is the number of 90-minute intervals elapsed (so $n = 1$ is the start).

(b) 6 hours = 4 intervals of 90 minutes, so $n = 5$ (after 4 intervals from the start).

$P = 400 \times 3^4 = 400 \times 81 = 32\,400$ bacteria.

(c) Solve $400 \times 3^{n-1} > 1\,000\,000$ :

$3^{n-1} > 2500$

$(n-1)\ln 3 > \ln 2500$

$n - 1 > \frac{\ln 2500}{\ln 3} = \frac{7.824}{1.099} = 7.12$

$n > 8.12$

So after $n = 9$ intervals, i.e., after 12 hours.

(d) $S_8 = \dfrac{400(3^8 - 1)}{3 - 1} = \dfrac{400 \times 6560}{2} = 1\,312\,000$ bacteria.

The total across 8 intervals is 1,312,000, which is well below the 5,000,000 capacity. The vessel is sufficient over this time period.

Question 3 — Logarithms in Exponential Growth Context [~6 marks]

The value $V$ (in dollars) of a vintage guitar is modelled by $V(t) = 1200 \times e^{0.065t}$ , where $t$ is the number of years since it was purchased.

(a) Find the initial value of the guitar.

(b) Find the value after 10 years, correct to the nearest dollar.

(d) The owner wants to sell when the guitar is worth at least $5000. Find the minimum whole number of years they must wait.

Show Solution

(a) $V(0) = 1200 \times e^0 = 1200$ (USD).

(b) $V(10) = 1200 \times e^{0.65} = 1200 \times 1.9155 = 2299$ (USD, nearest dollar).

(c) Set $V(t) = 2400$ (double the initial value):

$1200 \times e^{0.065t} = 2400$

$e^{0.065t} = 2$

$0.065t = \ln 2 = 0.6931$

$t = \frac{0.6931}{0.065} = 10.7 \text{ years}$

(d) Solve $1200 \times e^{0.065t} \ge 5000$ :

$e^{0.065t} \ge \frac{5000}{1200} = 4.1\overline{6}$

$0.065t \ge \ln(4.167) = 1.4271$

$t \ge \frac{1.4271}{0.065} = 21.96 \text{ years}$

The owner must wait at least 22 years.

GIVEN	z = r(cosθ + i sinθ) = r cis θ
GIVEN	z = re^iθ (Euler form)
MEMORISE	\|z\| = √(a² + b²)
MEMORISE	arg(z) — sketch point, use quadrant formula

GIVEN	z&sub1;z&sub2; = r&sub1;r&sub2; cis(θ&sub1; + θ&sub2;)
GIVEN	z&sub1;/z&sub2; = (r&sub1;/r&sub2;) cis(θ&sub1; − θ&sub2;)

GIVEN	(r cis θ)ⁿ = rⁿ cis(nθ)
MEMORISE	z + 1/z = 2cosθ (when \|z\|=1)
MEMORISE	z − 1/z = 2i sinθ (when \|z\|=1)

GIVEN	w^1/n = r^1/n cis((θ + 2πk)/n), k=0..n-1
MEMORISE	Sum of nth roots of unity = 0
MEMORISE	1 + ω + ω² = 0 (cube roots)

MEMORISE	z* = a − bi
MEMORISE	z · z* = \|z\|² (always real)
MEMORISE	z + z* = 2Re(z)
MEMORISE	z − z* = 2i Im(z)

MEMORISE	\|z − a\| = r → Circle, centre a, radius r
MEMORISE	\|z − a\| = \|z − b\| → Perpendicular bisector
MEMORISE	arg(z − a) = θ → Ray from a

MEMORISE	z² + az + b = 0: sum = −a, product = b
MEMORISE	Conjugate root theorem: real coeff → roots come in conjugate pairs

IB Math AI SL — Number and Algebra

Section 1: Scientific Notation, Rounding, and Estimation

1.1 Scientific Notation

1.2 Significant Figures and Rounding

Section 2: Arithmetic Sequences and Series

2.1 The nnnth Term

2.2 The Sum of nnn Terms

2.3 Applications

Section 3: Geometric Sequences and Series

3.1 The nnnth Term

3.2 The Sum of nnn Terms

3.3 Infinite Geometric Series

Section 4: Financial Mathematics

4.1 Compound Interest

4.2 Using the GDC Finance Solver (TVM)

4.3 Annuities and Amortization

Section 5: Logarithms in Context

5.1 The Logarithm as an Inverse

5.2 Solving Exponential Equations

5.3 Half-Life

Section 6: Practice Questions

Paper 1 Style (Short Answer)

Paper 2 Style (Extended Response)

May 2026 Prediction Questions

IB Formula Booklet — Complex Numbers

Modulus & Polar Form

Polar Multiplication & Division

De Moivre's Theorem

nth Roots

Conjugate & Arithmetic

Loci

Vieta's Formulas

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2.1 The $n$ th Term

2.2 The Sum of $n$ Terms

3.1 The $n$ th Term

3.2 The Sum of $n$ Terms