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Patterns & Sequences

Arithmetic, quadratic and geometric sequences and series, sigma notation, and both examinable proofs. Notes, past papers and a quiz — everything for this topic is one click away.

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Grade 12 CAPS Mathematics

Patterns, Sequences & Series

Arithmetic, quadratic and geometric sequences and series, sigma notation, and both examinable proofs.

Recognising the Three Types

Identify the type first — before writing any formula.

Arithmetic (linear)

First differences are constant. Subtract consecutive terms: \(d=T_n-T_{n-1}\).

Example: \(3;7;11;15;\ldots\) with \(d=4\).

Quadratic

Second differences are constant. Need at least 4 terms to confirm.

Example: \(6;12;22;36;\ldots\)

Geometric (exponential)

A constant ratio between terms. Divide consecutive terms: \(r=T_n\div T_{n-1}\).

Example: \(2;6;18;54;\ldots\) with \(r=3\).

Important
Arithmetic questions start with subtraction. Geometric questions start with division. Quadratic questions require first and second differences. Write your conclusion clearly — that method step often earns marks.

Arithmetic Sequence: Definition

Constant first difference.

Definition

A sequence where the difference between any two consecutive terms is constant — the common difference \(d\).

\[a,\;\;a+d,\;\;a+2d,\;\;a+3d,\;\ldots\]
Common difference
\[\boxed{d=T_n-T_{n-1}}\]

General Term: \(T_n=a+(n-1)d\)

On your formula sheet — and the derivation is worth knowing.

Formula
\[T_n=a+(n-1)d\]

\(a\) is the first term \((T_1)\), \(d\) is the common difference, \(n\) is the term position.

Show derivation
  1. 1\(T_1=a\). No \(d\) has been added yet.
  2. 2\(T_2=a+d\). One \(d\) has been added.
  3. 3\(T_3=a+2d\). Two \(d\)'s have been added.
  4. 4The coefficient of \(d\) is always one less than the term number, so \(T_n=a+(n-1)d\).
Worked example Level 1

Sequence \(4;10;16;\ldots\) Find \(T_n\), then \(T_{50}\).

Show solution
  1. 1\(a=4\), \(d=6\), so \(T_n=4+(n-1)(6)=\boxed{6n-2}\).
  2. 2\(T_{50}=6(50)-2=\boxed{298}\)
Worked example — find \(a\) and \(d\) from two terms Level 2-3

An arithmetic series has its 19th term equal to \(11\) and its 31st term equal to \(5\). Find \(a\) and \(d\).

Show solution
  1. 1Write the two term equations: \(T_{19}=a+18d=11\) and \(T_{31}=a+30d=5\).
  2. 2Subtract: \(12d=-6 \Rightarrow d=-\tfrac12\).
  3. 3Substitute back: \(a+18(-\tfrac12)=11 \Rightarrow a=20\).
  4. 4So the sequence starts \(\boxed{20;\,19.5;\,19;\ldots}\)

Sum Formula: \(S_n=\tfrac{n}{2}[2a+(n-1)d]\)

Two equivalent forms — and the proof is examinable.

Two forms, both on your formula sheet
\[S_n=\frac{n}{2}[2a+(n-1)d] \qquad\text{or}\qquad S_n=\frac{n}{2}(a+l)\]

Use \(S_n=\tfrac{n}{2}(a+l)\) only when the last term \(l\) is given or easy to find.

Show full proof (examinable)
  1. 1Write the series forward, \(l\) = last term: \(S_n=a+(a+d)+(a+2d)+\cdots+(l-d)+l\)
  2. 2Write the same series backward: \(S_n=l+(l-d)+(l-2d)+\cdots+(a+d)+a\)
  3. 3Add the two lines. There are \(n\) pairs, each adding to \((a+l)\): \(2S_n=n(a+l)\)
  4. 4Divide by 2: \(S_n=\tfrac{n}{2}(a+l)\). Substitute \(l=a+(n-1)d\): \(\boxed{S_n=\tfrac{n}{2}[2a+(n-1)d]}\)
Worked example — sum of 31 terms Level 1

Using \(a=20\), \(d=-\tfrac12\) (from the earlier example), find \(S_{31}\).

Show solution
  1. 1\(S_{31}=\dfrac{31}{2}\left[2(20)+30\left(-\tfrac12\right)\right]=\dfrac{31}{2}(25)=\boxed{387.5}\)
Worked example — sum of all negative terms Level 2-3

For \(3;-1;-5;\ldots-85;-89\), calculate the sum of all negative terms.

Show solution
  1. 1The negative terms run from \(-1\) to \(-89\) — 23 terms.
  2. 2\(S_{23}=\dfrac{23}{2}(-1+-89)=\dfrac{23}{2}(-90)=\boxed{-1\,035}\)
Worked example — series in context Level 3

Calculate the sum of all integers from 100 to 300 that are multiples of 4.

Show solution
  1. 1Series: \(100+104+108+\cdots+300\)
  2. 2Number of terms: \(300=100+(n-1)4 \Rightarrow n=51\)
  3. 3\(S_{51}=\dfrac{51}{2}(100+300)=\boxed{10\,200}\)
Worked example — find \(n\) from \(S_n\) Level 3

\(5+3+1+\cdots=-216\). Find the number of terms.

Show solution
  1. 1\(a=5\), \(d=-2\): \(-216=\tfrac{n}{2}[10+(n-1)(-2)]\)
  2. 2\(-432=12n-2n^2 \Rightarrow n^2-6n-216=0 \Rightarrow (n-18)(n+12)=0\)
  3. 3Reject the negative root: \(\boxed{n=18}\) terms.
Worked example — find \(T_n\) from \(S_n\) Level 1

Given \(S_n=3n^2-2n\), find \(T_9\).

Show solution
  1. 1\(S_9=3(81)-18=225\)
  2. 2\(S_8=3(64)-16=176\)
  3. 3\(T_9=S_9-S_8=225-176=\boxed{49}\)
Key identity
\(T_n=S_n-S_{n-1}\) for \(n\ge2\), and \(T_1=S_1\). This is how you recover a term when only the sum formula is given.
Quick Check

The common difference of \(5;9;13;17;\ldots\) is:

If \(a=3\) and \(d=5\), which formula is correct?

If \(S_n=3n^2-2n\), then \(T_4\) equals:

Quadratic Sequence

Constant second difference — the exam's Level 2-3 gateway topic.

Definition

Second differences are constant. The first differences themselves form an arithmetic sequence.

General term
\[T_n=an^2+bn+c\]
Important
You need at least 4 terms to confirm a quadratic sequence — with only 3 terms you get only one second difference, so you can't tell whether it will stay constant.
Position\(T_1\)\(T_2\)\(T_3\)\(T_4\)
Terms (\(13;27;45;67;\ldots\))13274567
First differences141822
Second differences44

Second differences are constant — quadratic. Next first difference is \(26\), so the next term is \(93\).

Finding \(T_n=an^2+bn+c\)

Three relationships, solved in order.

Three key relationships
\[2a=\text{second difference}\qquad 3a+b=T_2-T_1\qquad a+b+c=T_1\]

Solve in order: find \(a\) first, then \(b\), then \(c\).

Worked example — shortcut method Level 2

For \(13;27;45;67;\ldots\), the second difference is 4.

Show solution
  1. 1\(2a=4 \Rightarrow a=2\)
  2. 2\(3a+b=T_2-T_1=14 \Rightarrow 6+b=14 \Rightarrow b=8\)
  3. 3\(a+b+c=T_1=13 \Rightarrow 2+8+c=13 \Rightarrow c=3\)
  4. 4So \(\boxed{T_n=2n^2+8n+3}\). Check \(n=3\): \(18+24+3=45\) ✓
Worked example — solve for \(n\) Level 1

Use \(T_n=2n^2-2n+3\). Which term equals \(4\,903\)?

Show solution
  1. 1\(2n^2-2n+3=4\,903 \Rightarrow 2n^2-2n-4\,900=0 \Rightarrow n^2-n-2\,450=0\)
  2. 2Factorise: \((n-50)(n+49)=0\)
  3. 3Reject the negative root: \(4\,903\) is the \(\boxed{50\text{th term}}\).
Quick Check

A quadratic sequence has constant:

If the second difference is 6, then \(a\) equals:

Which rule matches \(13;27;45;67;\ldots\)?

Geometric Sequence

Constant ratio — models growth and decay by a fixed factor.

Definition

Each term is obtained by multiplying the previous term by a constant ratio \(r\).

\[a,\;\;ar,\;\;ar^2,\;\;ar^3,\;\ldots\]
Common ratio
\[\boxed{r=\dfrac{T_n}{T_{n-1}}}\]
Why this matters
Geometric thinking appears when a quantity changes by the same factor, not the same amount — doubling growth, repeated discounting, or halving over time. Convergence matters once \(|r|<1\): the terms shrink toward zero, and a sum to infinity becomes possible.

General Term and Finite Sum

The sum-formula proof is examinable, same as the arithmetic one.

Formulas, on your formula sheet
\[T_n=ar^{n-1}\]\[S_n=\frac{a(r^n-1)}{r-1}\quad\text{or}\quad S_n=\frac{a(1-r^n)}{1-r}\]

Both finite-sum forms are equivalent when \(r\neq1\).

Show full proof (examinable)
  1. 1Write the series: \(S_n=a+ar+ar^2+\cdots+ar^{n-1}\)
  2. 2Multiply by \(r\): \(rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n\)
  3. 3Subtract the first line from the second — the middle terms cancel: \(rS_n-S_n=ar^n-a \Rightarrow S_n(r-1)=a(r^n-1)\)
  4. 4Divide by \(r-1\): \(\boxed{S_n=\dfrac{a(r^n-1)}{r-1}}\)
Worked example — direct substitution Level 1

A geometric sequence has \(a=3\) and \(r=2\). Find \(T_6\).

Show solution
  1. 1\(T_n=ar^{n-1}\), so \(T_6=3(2)^{6-1}=3(2)^5\)
  2. 2\(T_6=3(32)=\boxed{96}\)
Worked example — find \(a\) and \(r\) from two terms Level 2-3

The 4th term of a geometric sequence is \(6\), and the 9th term is \(0.1875\). Determine the sequence.

Show solution
  1. 1\(T_9=ar^8\), \(T_4=ar^3\). Dividing: \(r^5=\dfrac{0.1875}{6}=\dfrac{1}{32}\)
  2. 2So \(r=\dfrac12\).
  3. 3\(6=a\left(\dfrac12\right)^3 \Rightarrow a=48\)
  4. 4The sequence is \(\boxed{48;24;12;6;\ldots}\)
Worked example — sum to infinity from a sourced question Level 1

The first two terms of a converging geometric series are \(8\) and \(m\), and the sum to infinity is \(12\). Determine the constant ratio.

Show solution
  1. 1\(12=\dfrac{8}{1-r} \Rightarrow 12(1-r)=8 \Rightarrow \boxed{r=\tfrac13}\)
Worked example — find \(n\) from \(S_n\) Level 2-3

\(2+6+18+\cdots=728\). Find \(n\).

Show solution
  1. 1\(a=2\), \(r=3\): \(728=\dfrac{2(3^n-1)}{2}=3^n-1\)
  2. 2\(3^n=729=3^6 \Rightarrow \boxed{n=6}\)

Convergence and Sum to Infinity

Only when \(|r|<1\).

Convergent series

Terms shrink toward zero, so the series adds to a finite value. Happens when \(|r|<1\).

Example: \(8;4;2;1;\ldots\) with \(r=\tfrac12\)

Divergent series

Terms grow or keep oscillating — no finite sum to infinity.

Example: \(2;4;8;16;\ldots\) with \(r=2\)

Sum to infinity, only when \(|r|<1\)
\[S_\infty=\frac{a}{1-r}\]
Worked example — standard sum to infinity Level 1

Sequence \(8;4;2;1;\ldots\), so \(a=8\), \(r=\tfrac12\).

Show solution
  1. 1Since \(\left|\tfrac12\right|<1\), the series converges.
  2. 2\(S_\infty=\dfrac{8}{1-\frac12}=\dfrac{8}{\frac12}=\boxed{16}\)
Worked example — for which \(x\) does it converge? Level 3-4

\(1+\dfrac{2x-5}{2}+\left(\dfrac{2x-5}{2}\right)^2+\cdots\) is convergent.

Show solution
  1. 1\(r=\dfrac{2x-5}{2}\). For convergence, \(-1<\dfrac{2x-5}{2}<1\)
  2. 2Multiply by 2: \(-2<2x-5<2\). Add 5: \(3<2x<7\)
  3. 3Divide by 2: \(\boxed{\tfrac32
Quick Check

The common ratio of \(48;24;12;6;\ldots\) is:

Which condition is correct for a sum to infinity to exist?

If \(2+6+18+\cdots=728\), then \(n\) equals:

Sigma Notation

Reading, interpreting, counting, and evaluating.

The symbol \(\Sigma\)

The Greek capital sigma means “add all these terms.”

The general term

The expression after \(\Sigma\). This is the term rule \(T_k\).

Lower limit

The value below \(\Sigma\) — where to start substituting.

Upper limit

The value above \(\Sigma\) — the last substitution value.

Important
The expression after \(\Sigma\) is the general term, not the sum formula. Substitute the starting value to find the first term, decide arithmetic or geometric, count the terms, then apply the correct sum formula.

Three-Step Method for Every Sigma Question

The same three steps, every time.

StepWhat to do
1Determine the type of series. Substitute the first two or three values; decide arithmetic, geometric, or neither.
2Count the number of terms: \(\text{upper limit}-\text{lower limit}+1\). This becomes \(n\).
3Apply the correct sum formula: arithmetic, geometric, or sum to infinity if it converges.
Worked example — direct evaluation Level 1

Evaluate \(\displaystyle\sum_{k=1}^{4}(2k+1)\).

Show solution
  1. 1Substitute \(k=1,2,3,4\) into \(2k+1\): \(3,\,5,\,7,\,9\)
  2. 2Add them: \(3+5+7+9=\boxed{24}\)
Worked example — arithmetic sigma Level 3

Given \(\displaystyle\sum_{n=1}^{m}(4n-19)=1189\), find \(m\).

Show solution
  1. 1First term: \(4(1)-19=-15\), common difference \(4\).
  2. 2\(\dfrac{m}{2}[2(-15)+(m-1)4]=1189\)
  3. 3Simplify: \(m(2m-17)=1189 \Rightarrow 2m^2-17m-1189=0\)
  4. 4Factorise: \((2m+41)(m-29)=0 \Rightarrow \boxed{m=29}\)
Worked example — geometric sigma with unknown upper limit Level 3

Find \(n\) if \(\displaystyle\frac{5}{3}\sum_{k=1}^{n}3^{k-1}=\frac{1820}{3}\).

Show solution
  1. 1The inner series is geometric, \(a=1\), \(r=3\): \(\sum_{k=1}^{n}3^{k-1}=\dfrac{3^n-1}{2}\).
  2. 2Substitute: \(\dfrac{5}{3}\cdot\dfrac{3^n-1}{2}=\dfrac{1820}{3}\)
  3. 3\(5(3^n-1)=3640 \Rightarrow 3^n=729=3^6 \Rightarrow \boxed{n=6}\)
Worked example — writing a series in sigma notation Level 2

\(1+5+9+\cdots+21\)

Show solution
  1. 1\(a=1\), \(d=4\), so \(T_k=4k-3\).
  2. 2Number of terms: \(4n-3=21 \Rightarrow n=6\)
  3. 3\(\boxed{\displaystyle\sum_{k=1}^{6}(4k-3)}\)
Worked example — infinite sum from sigma Level 3

Calculate \(\displaystyle\sum_{p=1}^{\infty}8(4)^{1-p}\).

Show solution
  1. 1\(T_1=8(4)^0=8\), \(T_2=8(4)^{-1}=2\), so \(r=\tfrac{2}{8}=\tfrac14\).
  2. 2\(\left|\tfrac14\right|<1\) — converges.
  3. 3\(S_\infty=\dfrac{8}{1-\frac14}=\boxed{\dfrac{32}{3}}\)
Quick Check

How many terms are in \(\sum_{k=3}^{10}\)?

The expression after \(\Sigma\) represents the:

\(\sum_{n=1}^{5}(3n-2)\) generates which kind of series?

Five-Step Strategy for Every Sequences Question

Use this method every time.

1. Identify the type. Check differences or ratio and write your conclusion clearly.
2. State the known values. Write \(a=\ldots\), \(d=\ldots\) or \(r=\ldots\), and \(n=\ldots\), before substituting.
3. Write the full formula. This protects your method marks.
4. Substitute carefully. Watch the brackets around \((n-1)\) and any negative values.
5. Verify your result. Reject a negative \(n\) — it must be a positive natural number.
Question typeFirst move
Identify the pattern typeCheck first differences, then second differences, then ratio if needed.
Find the \(n\)th termWrite down the known values first, then choose the right \(T_n\) formula.
Which term equals…?Set \(T_n\) equal to the given value and solve for \(n\).
Find a sumDecide arithmetic or geometric, identify \(a\), \(d\) or \(r\), and \(n\) or \(l\).
Given \(S_n\), find a termUse \(T_n=S_n-S_{n-1}\).
Sigma notationExpand the first few terms, count the terms, then choose the sum formula.
Convergence / sum to infinityWrite the ratio first, then solve \(|r|<1\) before using \(S_\infty\).

Cognitive Levels: What the Exam Tests

The same four DBE levels used throughout this site.

Level 1: Knowledge

Recall formulas, identify the sequence type, substitute directly, find the next term.

Level 2: Routine procedures

Find \(T_n\), \(S_n\), a specific term, or \(n\), using standard methods.

Level 3: Complex procedures

Multi-step questions, linked conditions, convergence questions with a variable.

Level 4: Problem solving

Unfamiliar contexts, combined ideas, interpreting patterns in practical situations.

Mixed end-of-topic drill Rising difficulty

Try each in order before revealing the answer.

Show all answers
  1. 1\(7;11;15;19;\ldots\) — arithmetic, next term \(\boxed{23}\).
  2. 2\(2;7;15;26;40;\ldots\) — \(\boxed{T_n=\dfrac{3n^2+n}{2}}\)
  3. 3\(48+24+12+\cdots\) — \(a=48\), \(r=\tfrac12\), so \(S_\infty=\boxed{96}\)
  4. 4\(\sum_{k=1}^{20}(3k-1)\) — arithmetic, first term 2, last term 59, 20 terms: \(\boxed{610}\)
  5. 5\(\sum_{k=1}^{\infty}(4x-1)^k\) converges when \(\boxed{0
  6. 6\(\sum_{k=1}^{n}2(3)^{k-1}=728 \Rightarrow \boxed{n=6}\). Not a sum-to-infinity question, since \(r=3\) means the series diverges.

Putting It All Together

Every Grade 12 sequences question is one of these decisions.

Ask yourselfThen
Finding one term?Use \(T_n\).
Finding a finite sum?Use \(S_n\).
Arithmetic or geometric?Difference vs. ratio.
Infinite geometric series?Check \(|r|<1\) first.
Given \(S_n\)?\(T_n=S_n-S_{n-1}\).
Two non-consecutive terms given?Form simultaneous equations.
Sigma notation?Interpret index, limits and the bracket structure first.
Both derivations are examinable
CAPS explicitly lists derivation of the arithmetic and geometric sum formulae as examinable content — make sure you can reproduce both proofs from this deck, not just apply the formulas.
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What CAPS Expects You to Know

The Grade 12 Number Patterns, Sequences and Series knowledge statement this page is built from.

  1. 1

    Number patterns, including arithmetic and geometric sequences and series.

  2. 2

    Sigma notation.

  3. 3

    Derivation and application of the formulae for the sum of arithmetic and geometric series. Derivation of the formulae is explicitly examinable.

How to Use This Lesson

A few practical notes before you start.

  • Identify the type first — before writing any formula. Arithmetic starts with subtraction, geometric with division, quadratic needs both first and second differences.
  • Both sum-formula proofs are examinable — don't just memorise the formulas, be able to reproduce the derivation.
  • For sigma notation, the expression after \(\Sigma\) is the general term, never the sum formula.
  • Try each worked example yourself before pressing “Show solution.”

Patterns, Sequences & Series, Concept by Concept

Four free, independent videos — not made by Equation Station SA.

Arithmetic Series

\(S_n=\tfrac{n}{2}[2a+(n-1)d]\) — where the sum formula comes from.

Khan Academy · Formula for arithmetic series

Quadratic Sequences

The reasoning behind \(2a=\)second difference — a Grade 11 skill you're expected to bring into Grade 12.

Beast Mode Maths · Why do we half the second difference?

Geometric Series & Convergence

\(S_\infty=\dfrac{a}{1-r}\), only when \(|r|<1\).

Khan Academy · Sum of an infinite geometric series

Sigma Notation

Reading the index, limits, and general term correctly.

Khan Academy · Sigma notation for sums

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Using \(T_n\) when the question asks for a sum

Read the question carefully — "term" means Tn, "sum" or "series" means Sn.

Forgetting the brackets in \((n-1)d\)

Writing \(a+nd\) instead of \(a+(n-1)d\) — always check against T1.

Using \(S_\infty\) without checking \(|r|<1\)

The sum to infinity formula is only valid for a convergent series.

Miscounting terms in sigma notation

Forgetting the \(+1\) in upper minus lower plus one.

Calling a pattern quadratic after one second difference

You need at least 4 terms to confirm a quadratic sequence, giving two second differences to compare.

Keeping a negative or non-integer \(n\)

A term position must be a positive natural number — always reject the other root.

Related Topics

Study these topics alongside Patterns & Sequences for a stronger foundation.

Algebra

Equations and inequalities

Study this

Finance & Growth

Interest and growth — the same compounding idea as geometric sequences

Study this

Functions & Graphs

Sketching and interpreting

Study this

Calculus

Differentiation and applications

Study this

Patterns & Sequences (Grade 11)

Quadratic number patterns, general term \(T_n=an^2+bn+c\)

Study this

Patterns & Sequences (Grade 10)

Linear number patterns, general term \(T_n=a+(n-1)d\)

Study this

Practise This Topic

You've done the notes above — now practise and test yourself.

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Past Papers
Patterns & Sequences Grade 12 Past Papers

Exam-style Grade 12 sequences and series questions arranged by cognitive level, with real citations from the DBE/provincial archive.

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Student Questions

Frequently Asked Questions

Straight answers to common Grade 12 CAPS questions about patterns, sequences, series, and the most important things to remember for Paper 1.

What are Patterns and Sequences in Grade 12 Maths?

Arithmetic, quadratic and geometric sequences, series, sigma notation and convergence; identify type, write Tn or Sn, then solve.

What should I remember first for Patterns and Sequences?

Check first/second differences or ratios to classify the sequence, then pick the relevant formula.

How do I improve at sequence questions?

Classify the pattern, write the general term explicitly, and practice converting between term and sum (Tn vs Sn).

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Past Question Papers, then finish with the Test Your Knowledge quiz as a self-check.

What mistakes should I avoid?

Common errors: using n instead of n-1, forgetting to test |r|<1 for infinite sums, and mixing up term vs sum questions.

How do Patterns and Sequences appear in CAPS exams?

Expect term rules, sums, sigma notation, and convergence checks — derivation of the sum formulae is explicitly examinable, often combined with reasoning steps.