Number and Algebra — IB Math AA HL AA HL Study Guide

For: IB Math AA HL candidates sitting IB Math: Analysis & Approaches HL.

Covers: All core Number and Algebra HL subtopics including arithmetic/geometric sequences and sigma notation, counting principles, logarithm laws and equations, complex numbers in Cartesian and polar form, HL-only mathematical induction, and solutions of systems of linear equations.

You should already know: IGCSE / pre-DP math, comfort with proof and algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AA HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Number and Algebra?

Number and Algebra is the foundational branch of mathematics focused on numerical patterns, symbolic manipulation, and abstract quantitative structures. It underpins every other topic in the IB Math AA HL syllabus, including calculus, statistics, and geometry, and appears in both calculator and non-calculator papers, often as a component of extended response (Section B) questions. It combines concrete numerical problem-solving with abstract proof skills that are assessed heavily in HL exams.

2. Sequences — arithmetic and geometric, sigma notation

Sequences are ordered lists of terms defined by a consistent rule. Arithmetic sequences have a constant common difference $d$ between consecutive terms, while geometric sequences have a constant common ratio $r$.

Key formulas

• Arithmetic nth term: $u_n=u_1+(n-1)d$, where $u_1$ is the first term
• Arithmetic sum of first $n$ terms: $S_n=\frac{n}{2}\left(2u_1+(n-1)d\right)=\frac{n}{2}\left(u_1+u_n\right)$
• Geometric nth term: $u_n=u_1{r}^{n-1}$
• Geometric sum of first $n$ terms: $S_n=u_1\frac{r^n-1}{r-1}$ for $r\ne 1$
• Infinite geometric sum: $S_{\infty }=\frac{u_1}{1-r}$, only valid if $|r|<1$ (convergent sequences)
• Sigma notation: ${\sum }_{k=a}^{b}f(k)$ denotes the sum of $f(k)$ for all integer values of $k$ from $a$ to $b$, with linearity properties: $\sum (kx+y)=k\sum x+\sum y$

Worked example: Find the sum of the first 8 terms of a geometric sequence with $u_1=3$ and $r=\frac{1}{3}$, and state the infinite sum if it exists. $$S_8=3\times \frac{{\left(\frac{1}{3}\right)}^8-1}{\frac{1}{3}-1}=3\times \frac{\frac{1}{6561}-1}{-\frac{2}{3}}=3\times \frac{-\frac{6560}{6561}}{-\frac{2}{3}}=\frac{3280}{729}\approx 4.5$$ Since $|r|=\frac{1}{3}<1$, $S_{\infty }=\frac{3}{1-\frac{1}{3}}=4.5$, which matches the approximate finite sum. Exam tip: Examiners regularly test infinite sum eligibility, so always check $|r|<1$ before applying the $S_{\infty }$ formula.

3. Counting — combinations, permutations, binomial theorem

Counting principles are used to calculate the number of possible outcomes for discrete events, a core skill for probability and combinatorics questions.

Key rules

• Permutations: Number of ways to arrange $r$ objects from $n$ distinct objects, where order matters: ${}^n{P}_r=\frac{n!}{(n-r)!}$
• Combinations: Number of ways to choose $r$ objects from $n$ distinct objects, where order does not matter: ${}^n{C}_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$
• Binomial theorem: The expansion of a binomial raised to a positive integer power $n$ is given by: $$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$$ The general (k+1)th term of the expansion is $T_{k+1}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)a^{n-k}{b}^k$

Worked example: Find the coefficient of $x^3$ in the expansion of $(3x-2{)}^5$. Use the general term formula, set $n=5$, $a=3x$, $b=-2$, and $k=2$ (since $n-k=3$ for the $x^3$ term): $$T_3=\left(\genfrac{}{}{0ex}{}{5}{2}\right)(3x{)}^3(-2{)}^2=10\times 27x^3\times 4=1080x^3$$ The coefficient of $x^3$ is 1080. Exam tip: Always use the general term to find specific coefficients instead of expanding the full binomial, as it saves time and reduces arithmetic errors in non-calculator papers.

4. Logarithms — laws and equations

Logarithms are the inverse operation of exponentiation, used to solve equations with unknown exponents and simplify exponential expressions.

Key definitions and laws

• If $b^c=a$, then ${\log }_{b}a=c$, where $b>0,b\ne 1,a>0$. Base $e$ logarithms are written as $\ln x$, base 10 as $\log x$.
• Product law: ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$
• Quotient law: ${\log }_b\left(\frac{x}{y}\right)={\log }_{b}x-{\log }_{b}y$
• Power law: ${\log }_b(x^k)=k{\log }_{b}x$
• Change of base law: ${\log }_{b}a=\frac{{\log }_{c}a}{{\log }_{c}b}$ for any positive $c\ne 1$

Worked example: Solve $5^{2x-1}=2^{x+2}$. Take natural logs of both sides and apply the power law: $$(2x-1)\ln 5=(x+2)\ln 2$$ Rearrange to isolate $x$: $$2x\ln 5-\ln 5=x\ln 2+2\ln 2$$ $$x(2\ln 5-\ln 2)=\ln 5+2\ln 2$$ $$x=\frac{\ln 5+2\ln 2}{2\ln 5-\ln 2}\approx \frac{1.61+1.39}{3.22-0.69}\approx 1.18$$ Exam tip: Always verify solutions to log equations by substituting back into the original expression, as log functions are only defined for positive arguments, so negative or zero solutions must be discarded.

5. Complex numbers — Cartesian and polar form

Complex numbers extend the real number system to include solutions for equations with negative discriminants, written in two standard forms.

Key properties

• Cartesian form: $z=a+bi$, where $a=\text{Re}(z)$ (real part), $b=\text{Im}(z)$ (imaginary part), and $i^2=-1$.
• Conjugate: $\overline{z}=a-bi$, used to simplify division of complex numbers: multiply numerator and denominator by the conjugate of the denominator to get a real denominator.
• Modulus: $|z|=\sqrt{a^2+b^2}$, the distance of $z$ from the origin on an Argand diagram.
• Polar form: $z=r(\cos \theta +i\sin \theta )=r\text{cis}\theta =r{e}^{i\theta }$, where $r=|z|$ and $\theta =\arg (z)$ (angle from the positive real axis, typically in the range $-\pi <\theta \le \pi$)
• De Moivre's theorem: For any integer $n$, $(r\text{cis}\theta {)}^n=r^n\text{cis}(n\theta )$, used to calculate powers and roots of complex numbers.

Worked example: Find the square roots of $z=-4i$. First convert $z$ to polar form: $r=\sqrt{0^2+(-4{)}^2}=4$, $\arg (z)=-\frac{\pi }{2}$, so $z=4\text{cis}\left(-\frac{\pi }{2}\right)$. Apply De Moivre's theorem for roots, adding $2\pi k$ to the argument for $k=0,1$: $$z^{\frac{1}{2}}=4^{\frac{1}{2}}\text{cis}\left(\frac{-\frac{\pi }{2}+2\pi k}{2}\right)=2\text{cis}\left(-\frac{\pi }{4}+\pi k\right)$$ For $k=0$: $2\text{cis}\left(-\frac{\pi }{4}\right)=\sqrt{2}-i\sqrt{2}$, for $k=1$: $2\text{cis}\left(\frac{3\pi }{4}\right)=-\sqrt{2}+i\sqrt{2}$ Exam tip: Always plot complex numbers on an Argand diagram before calculating the argument to avoid quadrant errors: for example, $z=-1+i$ has $\arg (z)=\frac{3\pi }{4}$, not $-\frac{\pi }{4}$.

6. Mathematical induction (HL only)

Mathematical induction is a formal proof technique used to verify statements about all positive integers $n$, assessed exclusively in HL papers.

Standard proof structure

1. Base case: Prove the statement holds for the smallest valid value of $n$ (usually $n=1$)
2. Inductive hypothesis: Assume the statement holds for some positive integer $n=k$, where $k\ge$ the base case value
3. Inductive step: Use the inductive hypothesis to prove the statement holds for $n=k+1$
4. Conclusion: State that by the Principle of Mathematical Induction (PMI), the statement holds for all $n\ge$ the base case value.

Worked example: Prove that for all positive integers $n$, ${\sum }_{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6}$.

• Base case $n=1$: LHS = $1^2=1$, RHS = $\frac{1\times 2\times 3}{6}=1$, so true for $n=1$.
• Inductive hypothesis: Assume ${\sum }_{r=1}^k{r}^2=\frac{k(k+1)(2k+1)}{6}$
• Inductive step: ${\sum }_{r=1}^{k+1}{r}^2={\sum }_{r=1}^k{r}^2+(k+1{)}^2=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$ Factor out $(k+1)$: $$=(k+1)\left(\frac{k(2k+1)+6(k+1)}{6}\right)=(k+1)\left(\frac{2k^2+7k+6}{6}\right)=\frac{(k+1)(k+2)(2k+3)}{6}$$ This matches the formula for $n=k+1$.
• Conclusion: By PMI, the statement holds for all positive integers $n$. Exam tip: The conclusion is always worth 1 mark, so never skip it even if the rest of the proof is complete.

7. Solutions of systems of linear equations

You will be assessed on solving systems of 3 linear equations with 3 variables, which represent intersections of 3 planes in 3D space. There are three possible solution cases:

1. Unique solution: All three planes intersect at a single point, the determinant of the coefficient matrix is non-zero, solvable via Gaussian elimination or inverse matrix methods.
2. No solution: The system is inconsistent (e.g., $0x+0y+0z=5$), planes are parallel or form a triangular prism with no common intersection.
3. Infinitely many solutions: The system is consistent but has fewer unique equations than variables, solutions are expressed in parametric form with one or more free variables, planes intersect along a line or are fully coincident.

Worked example: Classify the solution type for the system: $$x+2y-z=3$$ $$2x-y+z=1$$ $$3x+y=5$$ Add the first two equations: $3x+y=4$, but the third equation is $3x+y=5$, so we get $0=1$, an inconsistent statement. There are no solutions. Exam tip: For non-calculator papers, Gaussian elimination is faster than inverse matrix methods, and you can check row operations as you go to avoid arithmetic errors.

8. Common Pitfalls (and how to avoid them)

• Pitfall: Using the infinite geometric sum formula when $|r|\ge 1$. Why: Students forget the convergence condition. Correct move: Always check $|r|<1$ before applying $S_{\infty }$; if not, state the infinite sum does not exist.
• Pitfall: Mixing up permutations and combinations in counting questions. Why: Students ignore the order of selection. Correct move: Ask "does swapping two selected items change the outcome?" If yes, use permutations; if no, use combinations.
• Pitfall: Retaining invalid solutions to log equations. Why: Students apply log laws without checking domain restrictions. Correct move: Substitute all solutions back into the original equation to confirm all log arguments are positive, discard any invalid results.
• Pitfall: Miscalculating complex number arguments for non-first quadrant points. Why: Students rely solely on $\arctan \left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)$ without checking signs. Correct move: Plot the point on an Argand diagram first to identify the quadrant, then adjust the arctan result accordingly.
• Pitfall: Skipping steps in induction proofs. Why: Students assume the inductive step is self-explanatory. Correct move: Explicitly reference the inductive hypothesis in the inductive step, and write the full conclusion referencing the PMI.

9. Practice Questions (IB Math AA HL Style)

Question 1

An arithmetic sequence has first term $u_1=5$ and 10th term $u_{10}=41$. A geometric sequence also has first term 5, and its 3rd term equals the 3rd term of the arithmetic sequence. Find the sum of the first 7 terms of the geometric sequence, and the infinite sum if it exists.

Solution

First find the arithmetic sequence common difference: $u_{10}=5+9d=41⟹d=4$. The 3rd arithmetic term is $5+2\times 4=13$. For the geometric sequence, $u_3=5r^2=13⟹r^2=\frac{13}{5}⟹r=\pm \sqrt{\frac{13}{5}}$. The absolute value of both ratios is $\sqrt{2.6}\approx 1.61>1$, so no infinite sum exists. Sum of first 7 terms: $S_7=5\times \frac{(\frac{13}{5}{)}^3-1}{\sqrt{\frac{13}{5}}-1}\approx 425.8$ (or exact form $\frac{25198}{125}$ for integer powers).

Question 2

Let $f(x)=\ln (2x+1)$ for $x>-\frac{1}{2}$. Use mathematical induction to prove that, for all positive integers $n$, $$ f^{(n)}(x) = \frac{(-1)^{,n-1},(n-1)!,2^{n}}{(2x+1)^{n}}. $$

Solution

• Base case $n=1$: $f^{\prime }(x)=\frac{2}{2x+1}$. The formula gives $\frac{(-1{)}^{0}0!2^1}{(2x+1{)}^1}=\frac{2}{2x+1}$. So true for $n=1$.
• Inductive hypothesis: assume the formula holds for $n=k$, i.e. $f^{(k)}(x)=(-1{)}^{k-1}(k-1)!2^k(2x+1{)}^{-k}$.
• Inductive step: differentiate both sides w.r.t. $x$:$$\begin{array}{rl}{f}^{(k+1)}(x)& =(-1{)}^{k-1}(k-1)!2^k\cdot \frac{d}{dx}\left[(2x+1{)}^{-k}\right]\\ & =(-1{)}^{k-1}(k-1)!2^k\cdot (-k)(2x+1{)}^{-k-1}\cdot 2\\ & =(-1{)}^{k}k!2^{k+1}(2x+1{)}^{-(k+1)}\\ & =\frac{(-1{)}^{(k+1)-1}((k+1)-1)!2^{k+1}}{(2x+1{)}^{k+1}}.\end{array}$$ This is the formula with $n=k+1$, so the result holds for $n=k+1$ whenever it holds for $n=k$.
• Conclusion: by mathematical induction, the formula holds for all positive integers $n$.

Question 3

Solve the system of equations, giving parametric solutions if infinitely many exist: $$x+y+z=6$$ $$2x-y+z=3$$ $$3x+0y+2z=9$$

Solution

Add first two equations: $3x+2z=9$, which matches the third equation, so there is one free variable. Let $z=t$, then $3x=9-2t⟹x=\frac{9-2t}{3}$. Substitute into first equation: $y=6-x-t=6-\frac{9-2t}{3}-t=\frac{18-9+2t-3t}{3}=\frac{9-t}{3}$. Solutions are $x=3-\frac{2t}{3},y=3-\frac{t}{3},z=t,t\in \mathbb{R}$.

10. Quick Reference Cheatsheet

11. What's Next

Number and Algebra is the backbone of all IB Math AA HL content, so mastery of these concepts will directly improve your performance in every other topic. Sequences and series are used to build Taylor and Maclaurin series in calculus, counting principles are foundational for probability and statistics, complex numbers are applied to polar coordinates and differential equations, induction is used to prove statements across all math domains, and linear systems are core to linear algebra concepts tested in Paper 3. You will regularly see these concepts combined with other topics in Section B extended response questions, so revising this content regularly will reduce the time you spend learning more advanced topics later in the course. If you have any questions about specific subtopics, worked examples, or exam strategies for Number and Algebra, you can ask Ollie, our AI tutor, at any time for personalized explanations and extra practice problems. You can also browse more study guides for IB Math AA HL on the homepage to build your knowledge of related topics and prepare for your exams.