Formula Sheet Overview

Number and Algebra is one of the most compact but far-reaching parts of IB Mathematics Analysis and Approaches. The formulae appear early in the course, yet they keep returning in modelling, functions, calculus and probability. A sequence problem can become a finance problem. A logarithm problem can become an exponential model. A binomial expansion can support algebraic approximation, coefficient comparison, or probability-style counting. That is why this page is written as a working guide rather than a bare list of equations.

The core shared formulae for AA SL and AA HL can be grouped into five families. First, arithmetic sequences and arithmetic series describe a quantity that changes by a constant difference. Second, geometric sequences and geometric series describe a quantity that changes by a constant ratio. Third, compound interest applies geometric growth to finance. Fourth, exponentials and logarithms provide the algebraic language for growth, decay and inverse operations. Fifth, the binomial theorem expands powers of a binomial and gives a systematic way to find specific terms or coefficients.

Use this page alongside the wider IB Mathematics resources and the course-specific IB Mathematics Analysis and Approaches overview. If you are building foundations before this chapter, the prior learning formulae for AA SL and AA HL are the best place to check algebraic manipulation, indices and basic functions. If you are an HL student, return to the shared formulae here first, then extend your revision with the Numbers and Algebra Formulae AA HL only page.

Where Number and Algebra Fits in IB Mathematics AA

IB Mathematics Analysis and Approaches is built for students who need strong algebraic technique, symbolic reasoning and mathematical proof habits. Number and Algebra is therefore not a separate island; it is the toolkit that lets later topics work. When you study Functions Formulae AA SL and AA HL, exponential and logarithmic functions depend directly on the laws in this guide. When you revise Calculus Formulae AA SL and AA HL, geometric series and binomial expansions can support limits, approximations and curve analysis. Even topics that look visual, such as Geometry and Trigonometry Formulae AA SL and AA HL, still require algebraic rearrangement.

The shared AA SL and AA HL content should be mastered by every Analysis and Approaches student. HL students then meet additional depth, but the shared formulae remain examinable and often appear inside longer questions. A common HL mistake is to jump into harder material while losing marks on the setup of a geometric sum, a logarithmic rearrangement, or a binomial coefficient. The safest preparation plan is to make this page automatic first, then move to HL-only extension work.

Students taking Applications and Interpretation use some similar algebraic tools, but their course emphasis is different. If you are comparing course pathways or supporting students in both courses, the Number and Algebra Formulae AI SL and AI HL page is the more suitable companion for AI students. This AA page keeps the focus on the formulae and reasoning style expected in Analysis and Approaches, where exact form, algebraic clarity and method are usually more important than a purely numerical answer.

The best way to learn this topic is to treat every formula as a statement about structure. Do not memorise \(u_n=u_1+(n-1)d\) as a string of symbols only. Understand that the \(n\)th term is the first term plus \(n-1\) jumps. Do not memorise \(S_\infty=\frac{u_1}{1-r}\) without the condition \(|r|<1\). Understand that the terms must shrink toward zero for an infinite total to settle. Do not quote \(\log_a(xy)=\log_a x+\log_a y\) without remembering that \(x>0\), \(y>0\), \(a>0\), and \(a\ne1\).

Arithmetic Sequences and Series

An arithmetic sequence is a list of terms in which each term is obtained by adding the same number to the previous term. That number is the common difference, written \(d\). If the sequence is \(4, 9, 14, 19, \ldots\), then \(u_1=4\) and \(d=5\). The \(n\)th term formula is based on a simple counting idea: the first term needs no jump, the second term uses one jump, the third term uses two jumps, and the \(n\)th term uses \(n-1\) jumps.

This formula is often the quickest route when a question asks for a particular term, asks which term has a given value, or gives two terms and asks for the first term or common difference. For example, if \(u_4=13\) and \(u_{10}=37\), subtracting the equations gives \(6d=24\), so \(d=4\). Then \(u_4=u_1+3d\) gives \(13=u_1+12\), so \(u_1=1\). The sequence is therefore \(1,5,9,13,\ldots\).

The sum of the first \(n\) terms of an arithmetic sequence is called an arithmetic series. The formula has two common forms. Use the first form when you know \(u_1\), \(d\), and \(n\). Use the second form when you know the first and last terms.

The reason the sum formula works is useful to know. If the series is written forward and backward, each paired column has the same total: \(u_1+u_n\), \(u_2+u_{n-1}\), \(u_3+u_{n-2}\), and so on. There are \(n\) terms, so the total of the two written lines is \(n(u_1+u_n)\). One line is half of that, giving \(S_n=\frac{n}{2}(u_1+u_n)\). This reasoning helps you reconstruct the formula if you forget it.

Arithmetic sequences appear in exam questions about salaries increasing by a fixed amount, rows of seats increasing by a fixed number, savings plans with a fixed monthly addition, staircases, tile patterns, and linear modelling. The phrase "increases by the same amount" is usually the signal. If the wording says "increases by the same percentage" or "multiplied by the same factor", it is usually not arithmetic; it is geometric or exponential.

The link between arithmetic sequences and linear functions is also important. If \(u_n=u_1+(n-1)d\), then expanding gives \(u_n=dn+(u_1-d)\). This is a linear expression in \(n\). The gradient is \(d\), and the intercept is not usually \(u_1\) because the sequence starts at \(n=1\), not \(n=0\). This distinction explains why students sometimes make one-place errors when comparing sequence notation with graph notation.

For more practice focused only on this topic, the arithmetic sequence calculator is useful for checking answers, while the arithmetic sequence complete guide gives a slower explanation with additional examples. This page keeps arithmetic sequences in the broader AA Number and Algebra formula set, so it also shows how arithmetic ideas differ from geometric and exponential ones.

Arithmetic Sequence Example

A theatre has 18 seats in the first row. Each new row has 4 more seats than the row before it. Find the number of seats in row 25 and the total number of seats in the first 25 rows. This is arithmetic because the number of seats increases by the same amount each time. We have \(u_1=18\), \(d=4\), and \(n=25\).

The answer should be written with context: row 25 has 114 seats, and the first 25 rows contain 1650 seats. In IB questions, method marks often reward the formula setup, but final communication still matters. A bare number without units or context can be ambiguous.

Common Arithmetic Errors

• Using \(nd\) instead of \((n-1)d\) in the \(n\)th term formula.
• Substituting the last term into the common difference position in the sum formula.
• Forgetting that \(n\) must usually be a positive integer when asking for a term number.
• Using an arithmetic formula for a fixed percentage growth problem.
• Giving a numerical answer without explaining what it represents in the context.

Geometric Sequences and Series

A geometric sequence is a list of terms in which each term is obtained by multiplying the previous term by the same non-zero number. That multiplier is the common ratio, written \(r\). If the sequence is \(3, 6, 12, 24, \ldots\), then \(u_1=3\) and \(r=2\). If the sequence is \(80, 40, 20, 10, \ldots\), then \(u_1=80\) and \(r=\frac{1}{2}\). The ratio may be positive, negative, fractional, or greater than 1, but it must be consistent.

The exponent \(n-1\) appears for the same reason as the arithmetic formula: the first term has had no multiplications after it is chosen, the second term has had one multiplication by \(r\), and the \(n\)th term has had \(n-1\) multiplications. When solving for \(n\), logarithms may be needed because the unknown appears in an exponent. This is one reason geometric sequences and logarithms belong in the same formula guide.

The finite sum of a geometric series can be written in two equivalent forms. The first form below is often convenient when \(r>1\). The second is often convenient when \(|r|<1\), because it avoids a negative denominator. Both are algebraically the same when \(r\ne1\).

If \(r=1\), the sequence is constant and the geometric sum formula with \(r-1\) in the denominator cannot be used. In that special case every term is \(u_1\), so \(S_n=nu_1\). This is a small condition, but it is a good example of why formulae should be understood rather than copied mechanically.

The infinite geometric series formula is one of the most important conditional formulae in the AA course. It is not valid for every geometric series. It can be used only when \(|r|<1\), meaning \(-1

For example, \(10+5+2.5+1.25+\cdots\) has \(u_1=10\) and \(r=0.5\), so \(S_\infty=\frac{10}{1-0.5}=20\). The partial sums approach 20 but never need to equal it after a finite number of terms. By contrast, \(10+20+40+\cdots\) has \(r=2\). The terms grow, so an infinite finite sum is impossible. The formula must not be used.

Geometric sequences appear in compound interest, depreciation, population growth, radioactive decay, repeated percentage change, medication concentration, bouncing height, and infinite area problems. The phrase "multiplied by", "reduced by a percentage", "increases by \(p\%\)", or "retains a fraction" usually signals a geometric model. For a wider topic route, use RevisionTown's Sequences and Series resources after this formula page.

Geometric Sequence Example

A phone originally costs 900 dollars and loses 18% of its value each year. Estimate its value after 5 years. A loss of 18% means the phone retains 82% of its value each year, so \(r=0.82\). The original value is \(u_1=900\) if year 1 is the starting value, but for elapsed time modelling it is often clearer to write \(V=900(0.82)^t\), where \(t\) is years after purchase.

The phone is worth about 333.37 dollars after 5 years. The exact indexing depends on how the question defines the first term. IB questions usually specify whether the initial value is at \(t=0\), after one year, or at the first listed term. Always align the exponent with the timeline before calculating.

Geometric Series Example

A ball rebounds to 60% of its previous height. It is dropped from 3 metres. Find the total vertical distance travelled until it comes to rest. The first downward distance is 3 metres. Each rebound height is travelled twice, once upward and once downward. The rebound heights form the infinite geometric series \(3(0.6)+3(0.6)^2+3(0.6)^3+\cdots\).

The convergence condition is satisfied because \(|0.6|<1\). The total vertical distance is 12 metres. This example is a common exam pattern because it tests more than the formula: it tests whether the first drop is counted once and the rebound distances are counted twice.

Compound Interest and Financial Modelling

Compound interest is a practical application of geometric growth. If money grows by a fixed percentage in each compounding period, then the balance is multiplied by the same factor each time. This is why the compound interest formula has the same structure as a geometric sequence. The present value is the starting term, and the growth factor is \(1+\frac{r}{100k}\).

In this formula, \(FV\) is the future value, \(PV\) is the present value, \(r\) is the annual nominal interest rate written as a percentage, \(k\) is the number of compounding periods per year, and \(n\) is the number of years. If interest is compounded monthly, \(k=12\). If it is compounded quarterly, \(k=4\). If it is compounded annually, \(k=1\).

The expression \(kn\) counts the total number of compounding periods. The expression \(\frac{r}{100k}\) converts the annual percentage rate into a decimal rate per compounding period. For instance, a nominal annual rate of 6% compounded monthly gives a monthly rate of \(\frac{6}{100\cdot12}=0.005\). The monthly growth factor is \(1.005\). Over 3 years there are 36 monthly periods, so the multiplier is \(1.005^{36}\).

Compound interest questions often ask for the future value, the present value, the rate, the time, or a comparison between compounding schedules. Rearranging for time usually requires logarithms. If \(1500=1000(1.04)^n\), then \(1.5=1.04^n\), so \(n=\frac{\log(1.5)}{\log(1.04)}\). This connection is a strong reason to practise compound interest and logarithms together.

For calculator checking on finance-only examples, use the compound interest calculator. For a deeper explanation of the finance formula itself, the compound interest formula page and compound interest formula guide are useful supporting resources. On this page, compound interest is presented as part of the AA Number and Algebra formula family, so the focus is on the algebraic structure behind the model.

Compound Interest Example

A student invests 2000 dollars at a nominal annual interest rate of 5.4% compounded quarterly. Find the value after 7 years. Here \(PV=2000\), \(r=5.4\), \(k=4\), and \(n=7\). The quarterly rate is \(\frac{5.4}{100\cdot4}=0.0135\), and the number of quarters is \(28\).

The investment is worth about 2910.17 dollars after 7 years. If the question asks for interest earned, subtract the present value: \(2910.17-2000=910.17\). This distinction between future value and interest earned is another frequent source of avoidable errors.

Exponents and Logarithms

Exponentials and logarithms are inverse ideas. The exponential statement \(a^x=b\) says that the base \(a\) raised to the power \(x\) gives \(b\). The logarithmic statement \(\log_a b=x\) says that \(x\) is the power needed on base \(a\) to produce \(b\). The same relationship is being described in two different forms.

The base conditions matter: \(a>0\), \(a\ne1\), and \(b>0\). Logarithms with non-positive arguments are not defined in the real-number system used in this part of the course. When solving logarithmic equations, write the domain restrictions before manipulating the equation, then check the final answer in the original expression.

The essential logarithm laws are the product law, quotient law, and power law. They are valid when all logarithms involved are defined. The laws do not mean that logarithms distribute over addition or subtraction. For example, \(\log_a(x+y)\) is not equal to \(\log_a x+\log_a y\). That false step is one of the most common errors in algebra and calculus.

The change of base formula lets you calculate logarithms in bases that are not directly available on a calculator. It also supports algebraic transformations in proof-style questions. If \(a>0\), \(a\ne1\), \(b>0\), \(b\ne1\), and \(x>0\), then

In practice, \(b\) is often 10 or \(e\), depending on the calculator keys and the context. In IB Mathematics AA, exact logarithmic form is often accepted when the question does not require a decimal approximation. For example, the solution to \(3^x=20\) is \(x=\log_3 20=\frac{\ln 20}{\ln 3}\). Rounding should be done only at the final step unless the question instructs otherwise.

The RevisionTown Exponents and Logarithms page is a useful companion when you want more examples of index laws, log laws and exponential equations. This formula page keeps the explanation tied to the AA SL and AA HL formula set and shows how logarithms connect to sequences, compound interest and binomial expressions.

Solving Exponential Equations

To solve an exponential equation, isolate the exponential expression first, then apply logarithms. If the bases can be made the same, exact comparison of powers may be easier. For example, \(2^{x+1}=16\) can be written as \(2^{x+1}=2^4\), so \(x+1=4\), and \(x=3\). But \(2^{x+1}=17\) cannot be solved by matching simple powers, so logarithms are needed.

If a decimal is required, use the change of base formula: \(x=\frac{\ln17}{\ln2}-1\). The exact form is often clearer because it preserves the algebraic structure.

Solving Logarithmic Equations

To solve a logarithmic equation, start with restrictions. For example, in \(\log_2(x-1)+\log_2(x+3)=3\), the arguments require \(x-1>0\) and \(x+3>0\), so \(x>1\). Then combine the logarithms using the product law.

The solutions are \(x=-1\pm2\sqrt3\). Only \(x=-1+2\sqrt3\) satisfies \(x>1\), so the other root must be rejected. The domain check is not optional; it is part of the solution.

Binomial Theorem

The binomial theorem gives a systematic expansion of \((a+b)^n\), where \(n\) is a non-negative integer in this shared AA SL and AA HL context. It avoids repeated multiplication and makes it possible to find a specific term directly. The coefficients come from combinations, written \(\binom{n}{r}\), and match the rows of Pascal's triangle.

The general term is often the most useful part of the theorem. The \((r+1)\)th term is

The indexing is important: \(r=0\) gives the first term, \(r=1\) gives the second term, and \(r=n\) gives the final term. If a question asks for the term containing \(x^5\), you usually set up the power of \(x\) in the general term and solve for \(r\). Then substitute that value of \(r\) into the full term, including the coefficient and any constants.

For example, in \((2x-3)^6\), the general term is \(\binom{6}{r}(2x)^{6-r}(-3)^r\). The power of \(x\) is \(6-r\). If we want the term in \(x^4\), then \(6-r=4\), so \(r=2\). The term is \(\binom{6}{2}(2x)^4(-3)^2=15\cdot16x^4\cdot9=2160x^4\).

The sign pattern depends on the binomial. If the binomial is \((a-b)^n\), then the signs alternate because the second part is negative. If \(n\) is even, the last term is positive. If \(n\) is odd, the last term is negative. Many coefficient errors come from calculating \(\binom{n}{r}\) correctly but forgetting the negative sign attached to \(b^r\).

For additional examples, the Binomial Expansion page is a focused resource, while Binomial Theorem provides a theorem-centred explanation. This page treats the binomial theorem as one part of the AA Number and Algebra formula set, so the examples are designed to sit beside sequences, logarithms and compound interest.

Binomial Example

Find the coefficient of \(x^3\) in \((1-2x)^7\). The general term is \(\binom{7}{r}(1)^{7-r}(-2x)^r\). The power of \(x\) is \(r\), so for \(x^3\), use \(r=3\).

The coefficient is \(-280\). The negative sign is part of the answer because the third power of \(-2x\) is negative.

How to Choose the Correct Formula in Exam Questions

Formula selection is often the difference between a short solution and a long, uncertain one. In Number and Algebra, the wording of the question usually gives strong clues. If a value changes by the same amount each step, think arithmetic. If it changes by the same multiplier, ratio, or percentage, think geometric or exponential. If money grows through repeated interest periods, think compound interest. If the unknown is in an exponent, think logarithms. If a power of a two-term expression is being expanded, think binomial theorem.

Before substituting numbers, write down what each symbol represents. For sequences, identify \(u_1\), \(d\), \(r\), \(n\), \(u_n\), and \(S_n\). For finance, identify \(PV\), \(FV\), \(r\), \(k\), and \(n\). For logarithms, identify the base and the argument, then state restrictions. For binomial expansions, identify \(a\), \(b\), \(n\), and whether you need the whole expansion or a particular term. This habit reduces errors because it separates interpretation from calculation.

Questions also differ in whether they require exact form or decimal form. If a solution involves \(\log_2 19\), exact form may be more appropriate unless the context asks for a rounded value. If a finance question asks for money, round to two decimal places only at the end. If a sequence question asks for the smallest integer \(n\) satisfying an inequality, a decimal answer from logarithms is not enough; you must interpret it as an integer term number.

IB Mathematics AA rewards clear mathematical communication. Even when technology is allowed, you should show the formula, substitution, and interpretation. A calculator can compute \(2000(1.0135)^{28}\), but it cannot show that you understood quarterly compounding. A calculator can find a logarithmic value, but it cannot state why the logarithm was needed. A formula page like this is most useful when it supports that written reasoning.

Worked Mixed Examples

Example 1: Finding the First Term and Difference

The 5th term of an arithmetic sequence is 22 and the 17th term is 70. Find the first term and the sum of the first 17 terms. Use \(u_n=u_1+(n-1)d\). The two pieces of information give \(u_1+4d=22\) and \(u_1+16d=70\). Subtracting gives \(12d=48\), so \(d=4\). Then \(u_1+16=22\), so \(u_1=6\).

The first term is 6, the common difference is 4, and the sum of the first 17 terms is 646.

Example 2: Geometric Sum with a Ratio Greater Than 1

A bacteria culture has 500 cells at the start of an observation. Each hour the number of cells is multiplied by 1.18. Estimate the total of the hourly recorded values for the first 10 hours, including the initial value. This is a finite geometric sum with \(u_1=500\), \(r=1.18\), and \(n=10\). Since the sum is finite, there is no requirement that \(|r|<1\).

The total of the 10 recorded values is approximately 21166 cells, depending on the rounding required. The infinite sum formula would be invalid because \(r=1.18>1\).

Example 3: Solving for Time in Compound Interest

An account contains 3000 dollars and grows at 4% per year compounded annually. How long will it take to reach 5000 dollars? We need \(5000=3000(1.04)^n\). Divide by 3000, then take logarithms.

If the account is checked continuously in a modelling sense, about 13.03 years is the answer. If interest is credited only at the end of each year and the question asks for the first whole year in which the balance is at least 5000 dollars, then 14 years is needed. The interpretation depends on the wording.

Example 4: Logarithmic Equation with Restrictions

Solve \(\log_3(x)+\log_3(x-2)=2\). The domain restrictions are \(x>0\) and \(x-2>0\), so \(x>2\). Combine the logarithms using the product law.

Only \(x=1+\sqrt{10}\) satisfies \(x>2\). The other solution is rejected because it makes the logarithm arguments invalid.

Example 5: Coefficient in a Binomial Expansion

Find the constant term in \(\left(2x-\frac{3}{x}\right)^8\). The general term is \(\binom{8}{r}(2x)^{8-r}\left(-\frac{3}{x}\right)^r\). The power of \(x\) is \((8-r)-r=8-2r\). For a constant term, the power of \(x\) must be 0, so \(8-2r=0\), giving \(r=4\).

The constant term is 90720. This example shows why the general term is more efficient than writing out the full expansion.

Connections Across the Formulae

The formulae on this page are connected by deeper ideas. Arithmetic sequences connect to linear change because equal inputs create equal output differences. Geometric sequences connect to exponential change because equal inputs create equal output ratios. Compound interest is a geometric process written in financial language. Logarithms solve exponential equations because they reverse the operation of exponentiation. The binomial theorem expands powers and reveals coefficient patterns that later support algebra, calculus and probability.

When revising, do not practise each formula only in isolation. Mix question types. For instance, a compound interest question may ask when an investment first exceeds a target, requiring logarithms and integer interpretation. A geometric sequence question may ask for a limiting sum, requiring the condition \(|r|<1\). A binomial expansion question may include negative or fractional terms, requiring careful exponent tracking. Mixed practice trains the decision-making that exam questions actually demand.

It is also useful to compare Number and Algebra with other formula areas. Function transformations and inverse functions build on exponentials and logarithms, so the functions formulae for AA SL and AA HL should feel more accessible once this page is secure. Calculus questions involving exponential models, rates of change, or approximations rely on the same algebraic language, so the calculus formulae for AA SL and AA HL are a natural next step. Probability questions involving repeated trials may use binomial coefficients, so the statistics and probability formulae for AA SL and AA HL also connect with the counting ideas here.

Exam Strategy for AA SL and AA HL Students

Start every Number and Algebra problem by identifying the type of change. Constant difference, constant ratio, repeated percentage, inverse exponential, or binomial power are different signals. Write a short line that defines your variables before using a formula. If the question gives context, keep units attached in your working. If the question asks for a term number, remember that term numbers are usually positive integers. If it asks for a value after a number of years, decide whether the initial value corresponds to \(t=0\) or \(n=1\).

For non-calculator work, simplify expressions before substituting. For calculator-allowed work, still write the exact setup first. IB marking often gives credit for the expression even if the final rounded value is slightly off. Conversely, a correct decimal with no setup may not show enough method in a structured response. The formulae are not only tools for getting answers; they are also part of the communication expected in the course.

Use exact notation where it improves clarity. Write \(\frac{1}{2}\) rather than 0.5 when the fraction is exact. Write \(\log_2 7\) or \(\frac{\ln 7}{\ln 2}\) when the question does not require a decimal. In finance questions, keep intermediate values unrounded and round the final money amount to the required precision. In binomial expansion, keep coefficients exact and simplify signs carefully.

HL students should be especially careful not to underestimate shared SL material. HL extension topics often appear in questions that begin with a shared technique. If the first part of a question asks for a geometric sum and the second part asks for a proof, approximation or limiting argument, a mistake in the first part can affect the whole problem. Solid shared formula knowledge protects marks across the paper.

Common Mistakes and How to Avoid Them

Practice Questions with Answers

Use these questions to check whether you can choose and apply the formulae without being told which one to use. Try each question before opening the answer. The explanations are intentionally concise so you can focus on method.

Revision Routine for This Formula Set

A strong revision routine should move from recognition to execution to interpretation. First, cover the formula names and test whether you can describe when each one is used without looking at the equations. Second, write each formula from memory and explain the meaning of every symbol. Third, complete short substitution questions. Fourth, complete mixed questions in which the formula type is not named. Finally, review errors and write one sentence explaining what cue you missed.

For sequences, practise deriving terms from patterns and deriving formulae from given terms. For series, practise deciding whether the question asks for a term or a sum. For compound interest, practise changing the compounding frequency and solving for unknown time. For logarithms, practise both converting forms and solving equations with restrictions. For binomial theorem, practise finding a particular term without expanding the entire expression.

Keep a small error log. Good entries are specific: "I used \(n\) instead of \(n-1\) in the arithmetic term formula," "I forgot that \(S_\infty\) needs \(|r|<1\)," or "I rejected no solutions after solving a logarithmic equation." Vague entries such as "I need to revise logs" are less useful because they do not identify the behaviour to change.

Frequently Asked Questions