Patterns, Sequences & Series: Grade 12 Maths Summary Notes

Q: How do I identify the type of sequence?

Subtract consecutive terms. If the first differences are constant, it is arithmetic. If not, subtract again. If the second differences are constant, it is quadratic. If dividing consecutive terms gives a constant ratio, it is geometric.

Q: When does a geometric series converge?

A geometric series converges only when the absolute value of the common ratio satisfies |r| < 1.

Q: How do I count terms in sigma notation?

Number of terms = upper limit - lower limit + 1. For example, the sum from k = 3 to k = 10 has 10 - 3 + 1 = 8 terms.

Arithmetic Sequence and Arithmetic Series

Constant first difference. Linear general term. Level 1 to 3.

Recognising the three types

Arithmetic sequence (linear)

First differences are constant.
Subtract consecutive terms.
\(d = T_n - T_{n-1}\)
The general term is linear.
Example: \(3;\;7;\;11;\;15;\ldots\) with \(d = 4\).

Quadratic sequence

Second differences are constant.
You need at least 4 terms to confirm it.
The first differences form an arithmetic sequence.
The general term is quadratic.
Example: \(6;\;12;\;22;\;36;\ldots\)

Geometric sequence (exponential)

There is a constant ratio between terms.
Divide consecutive terms.
\(r = T_n \div T_{n-1}\)
The general term is exponential.
Example: \(2;\;6;\;18;\;54;\ldots\) with \(r = 3\).

Important:

Identify the type first before writing any formula. Arithmetic sequence questions start with subtraction. Geometric sequence questions start with division. Quadratic sequence questions require first and second differences. Write your conclusion clearly because that method often earns marks.

Definition and common difference

Definition: Arithmetic sequence

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

General form: \(\;a,\quad a+d,\quad a+2d,\quad a+3d,\quad\ldots\)

Common difference

\[\boxed{d = T_n - T_{n-1}}\]

Exam warning: Negative common difference

The sequence \(3;\;-1;\;-5;\;\ldots\) from a September 2024 Mpumalanga pre-trial paper has \(d = -4\). It is still an arithmetic sequence because the difference is constant. Always check the sign of \(d\) before substituting.

Source: G12 Maths P1 Mpumalanga Pre-Trial Sept 2024, Q2.2

Worked example: Use two given terms to find \(a\) and \(d\) (Level 2 to 3)

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

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

Source: 2025 G12 March Khayelitsha PLC, Q1.1

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

Formula: On your formula sheet

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

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

Show derivation of \(T_n = a + (n-1)d\)

\(T_1 = a\). No \(d\) has been added yet.

\(T_2 = a + d\). One \(d\) has been added.

\(T_3 = a + 2d\). Two \(d\) values have been added.

The pattern shows that the coefficient of \(d\) is always one less than the term number, so \(T_n = a + (n-1)d\).

Worked example: Find \(T_n\), then \(T_{50}\) (Level 2)

Sequence: \(4;\;10;\;16;\;\ldots\), so \(a = 4\) and \(d = 6\).

1\(T_n = 4 + (n-1)(6) = 6n - 2\)
2\(T_{50} = 6(50) - 2 = \mathbf{298}\)

Source: G12 Maths P1 NW Sept 2024, Q2.2

Worked example: Find the sum of the first 31 terms (Level 3)

1From the previous example, \(a = 20\) and \(d = -\tfrac{1}{2}\).
2Use \(S_n = \dfrac{n}{2}[2a + (n-1)d]\).
3\(S_{31} = \dfrac{31}{2}[2(20) + 30(-\tfrac{1}{2})] = \dfrac{31}{2}(25) = \mathbf{387.5}\)

Source: 2025 G12 March Khayelitsha PLC, Q1.2

Worked example: Sum of an arithmetic series in context (Level 3)

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

1Write the arithmetic series: \(100 + 104 + 108 + \cdots + 300\).
2Find the number of terms: \(300 = 100 + (n-1)4 \Rightarrow 4n = 204 \Rightarrow n = 51\).
3Now sum: \(S_{51} = \dfrac{51}{2}(100 + 300) = \mathbf{10\,200}\).

Source: 2025 G12 SACAI P1 May_June, Q2.2

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

Two equivalent forms: Both are 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 \(S_n = \frac{n}{2}(a + l)\) only when the last term \(l\) is given or easy to find.

Full proof: Arithmetic series sum formula

Exam warning:

Do not use the formula you are proving inside the proof itself. Use only \(T_n = a + (n-1)d\) and the structure of the sum.

Write the series forward, where \(l\) is the last term: \[S_n = a + (a+d) + (a+2d) + \cdots + (l-d) + l\]

Write the same series backward: \[S_n = l + (l-d) + (l-2d) + \cdots + (a+d) + a\]

Add the two lines. There are \(n\) pairs, and each pair adds to \((a+l)\): \[2S_n = n(a+l)\]

Divide by 2: \[S_n = \tfrac{n}{2}(a+l)\] Then substitute \(l = a + (n-1)d\): \[S_n = \tfrac{n}{2}[2a + (n-1)d]\]

Worked example: Sum of all negative terms (Level 2 to 3)

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

\[S_{22} = \frac{22}{2}(-3 + -89) = 11(-92) = \mathbf{-1\,012}\]

Source: G12 Maths P1 Mpumalanga Pre-Trial Sept 2024, Q2.2.2

Worked example: Find \(n\) when \(S_n\) is given (Level 3)

\(5 + 3 + 1 + \cdots = -216\), so \(a = 5\) and \(d = -2\).

1\(-216 = \tfrac{n}{2}[2(5) + (n-1)(-2)]\)
2\(-432 = 12n - 2n^2 \Rightarrow 2n^2 - 12n - 432 = 0 \Rightarrow n^2 - 6n - 216 = 0\)
3\((n-18)(n+12) = 0 \Rightarrow n = 18\), since \(n\) must be a positive natural number.
4So 18 terms sum to \(-216\).

Worked example: Find \(T_n\) from \(S_n\) (Level 3)

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

1\(S_9 = 3(81) - 18 = 225\)
2\(S_8 = 3(64) - 16 = 176\)
3\(T_9 = S_9 - S_8 = 225 - 176 = \mathbf{49}\)

Key identity: \(\boldsymbol{T_n = S_n - S_{n-1}}\) for \(n \geq 2\), and \(T_1 = S_1\).

Common mistake:

The formula is \(a + (n-1)d\), not \(a + nd\). The bracket matters because the first term contains zero copies of the common difference.

Quick practice: Arithmetic sequence and arithmetic series

1. Given \(5;\;9;\;13;\;17;\;\ldots\), find the general term and then state which term equals 217.

2. The sum \(22 + 28 + 34 + \cdots\) equals 1870. Find the number of terms.

3. \(3x+1;\;2x;\;3x-7\) are the first three terms of an arithmetic sequence. Find \(x\).

4. \(S_n = 3n^2 - 2n\). Determine \(T_9\).

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Quadratic Sequence

Constant second difference. Quadratic general term. Level 2 to 3.

Definition and difference tables

Definition: Quadratic sequence

A sequence where the second differences are constant. The first differences form an arithmetic sequence.

Important:

You need at least 4 terms to confirm a quadratic sequence. With only 3 terms you get only one second difference, so you cannot tell whether the pattern will remain constant.

Worked example: Difference table (Level 2)

Sequence: \(13;\;27;\;45;\;67;\;\ldots\)

Position	\(T_1\)	\(T_2\)	\(T_3\)	\(T_4\)
Terms	13	27	45	67
First differences	14	18	22
Second differences	4	4

The second differences are constant, so this is a quadratic sequence.

Next first difference: \(26\), so the next term would be \(93\).

Source: 2024 G12 CT March Sedibeng West, Q2.1

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

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.

1\(2a = 4 \Rightarrow a = 2\)
2\(3a + b = T_2 - T_1 = 14 \Rightarrow 6 + b = 14 \Rightarrow b = 8\)
3\(a + b + c = T_1 = 13 \Rightarrow 2 + 8 + c = 13 \Rightarrow c = 3\)
4So \(T_n = 2n^2 + 8n + 3\). Checking \(n = 3\) gives \(45\), which matches the sequence.

Source: 2024 G12 CT March Sedibeng West, Q2.1.1

Worked example: Which term equals \(4\,903\)? (Level 3)

Use the quadratic rule \(T_n = 2n^2 - 2n + 3\).

1Set the term rule equal to the target value: \(2n^2 - 2n + 3 = 4\,903\).
2\(2n^2 - 2n - 4\,900 = 0 \Rightarrow n^2 - n - 2\,450 = 0\).
3Factorise: \((n - 50)(n + 49) = 0\).
4Reject the negative root and keep \(n = 50\).
5So \(4\,903\) is the 50th term.

Source: 2025 G12 Free State P1 June, Q2.1.3

Exam warning:

After finding the rule, substitute \(n = 1\), \(n = 2\), and \(n = 3\) back into \(T_n\). This short check can save the whole question if you made a sign error.

Quick practice: Quadratic sequence

1. Sequence: \(2;\;7;\;15;\;26;\;40;\;\ldots\). Find \(T_n\), then determine which term equals 260.

2. Find \(T_6\) and \(T_n\) for \(-1;\;3;\;9;\;17;\;27;\;\ldots\)

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Geometric Sequence and Geometric Series

Constant ratio. Exponential general term. Finite and infinite sums. Level 1 to 4.

Definition and common ratio

Definition: Geometric sequence

A sequence where each term is obtained by multiplying the previous term by a constant ratio \(r\).

General form: \(\;a,\quad ar,\quad ar^2,\quad ar^3,\quad\ldots\)

Common ratio

\[\boxed{r = \dfrac{T_n}{T_{n-1}}}\]

Worked example: Determine a sequence from two known terms (Level 2 to 3)

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

1Use \(T_9 = ar^8\) and \(T_4 = ar^3\). Dividing gives \(r^5 = \dfrac{0.1875}{6} = \dfrac{1}{32}\).
2So \(r = \dfrac{1}{2}\).
3Now \(6 = a\left(\dfrac{1}{2}\right)^3\), so \(a = 48\).
4The sequence is \(48;\;24;\;12;\;6;\;\ldots\)

Source: 2025 G12 SACAI P1 May_June, Q2.1

General term and finite sum

Formulas: On your formula sheet

\[T_n = ar^{n-1}\] \[S_n = \frac{a(r^n-1)}{r-1} \qquad \text{or} \qquad S_n = \frac{a(1-r^n)}{1-r}\]

Both finite sum forms are equivalent when \(r \neq 1\). Use the version that keeps your arithmetic cleaner.

Full proof: Geometric series sum formula

Write the geometric series: \[S_n = a + ar + ar^2 + \cdots + ar^{n-1}\]

Multiply the whole line by \(r\): \[rS_n = ar + ar^2 + \cdots + ar^{n-1} + ar^n\]

Subtract the first line from the second line. The middle terms cancel: \[rS_n - S_n = ar^n - a \Rightarrow S_n(r-1) = a(r^n - 1)\]

Divide by \(r-1\): \[S_n = \frac{a(r^n - 1)}{r - 1}\] This can also be written as \[S_n = \frac{a(1-r^n)}{1-r}\]

Worked example: Sum to infinity from a sourced question (Level 3)

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

\[12 = \frac{8}{1-r} \Rightarrow 12(1-r) = 8 \Rightarrow r = \mathbf{\frac{1}{3}}\]

Source: 2025 G12 SACAI P1 May_June, Q2.3

Worked example: Find \(n\) when \(S_n\) is given (Level 2 to 3)

\(2 + 6 + 18 + \cdots = 728\), so \(a = 2\) and \(r = 3\).

1\(728 = \dfrac{2(3^n - 1)}{2} = 3^n - 1\)
2So \(3^n = 729 = 3^6\), which means \(n = 6\).

Convergence and sum to infinity

Convergent series

The terms shrink toward zero, so the series adds to a finite value. This happens when \(|r| < 1\).

Example: \(8;\;4;\;2;\;1;\;\ldots\) with \(r = \tfrac{1}{2}\)

Divergent series

The terms grow or keep oscillating, so there is 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 2)

Sequence: \(8;\;4;\;2;\;1;\;\ldots\), so \(a = 8\) and \(r = \tfrac{1}{2}\).

Since \(\left|\tfrac{1}{2}\right| < 1\), the series converges.

\[S_\infty = \frac{8}{1-\frac{1}{2}} = \frac{8}{\frac{1}{2}} = \mathbf{16}\]

Worked example: For which \(x\) does the series converge? (Level 3 to 4)

The geometric series below is convergent: \(1 + \dfrac{2x-5}{2} + \left(\dfrac{2x-5}{2}\right)^2 + \cdots\)

1The common ratio is \(r = \dfrac{2x-5}{2}\).
2For convergence, \(-1 < \dfrac{2x-5}{2} < 1\).
3Multiply through by \(2\): \(-2 < 2x-5 < 2\).
4Add \(5\): \(3 < 2x < 7\).
5Divide by \(2\): \(\dfrac{3}{2} < x < \dfrac{7}{2}\).

Source: 2024 G12 SACAI P1 May_June, Q2.3.1

Exam warning:

State and verify \(|r| < 1\) before using \(S_\infty\). That line is usually one of the method marks in this type of question.

Quick practice: Geometric sequence and geometric series

1. Find the 10th term of \(3;\;6;\;12;\;\ldots\)

2. The first two terms are \((x+3)\) and \((x^2-9)\). For which \(x\) does the series converge?

3. The first four terms are \(7;\;x;\;y;\;189\) with \(r = 3\). Find \(x\) and \(y\).

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Sigma Notation

Reading, interpreting, counting, and evaluating. Level 2 to 4.

Anatomy of a sigma expression

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 \(\Sigma\). It tells you where to start substituting.

Upper limit

The value above \(\Sigma\). It tells you the last substitution value.

Common mistake: Counting terms

\[\text{Number of terms} = \text{upper limit} - \text{lower limit} + 1\]

\(\displaystyle\sum_{k=3}^{10}\) has \(10 - 3 + 1 = \mathbf{8}\) terms.

\(\displaystyle\sum_{k=0}^{n}\) has \(n + 1\) terms.

\(\displaystyle\sum_{k=5}^{n}\) has \(n - 4\) terms.

Important:

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

Three-step method for every sigma question

Determine the type of series

Substitute the first two or three values. Decide whether the series is arithmetic, geometric, or neither.

Count the number of terms

\(\text{Number of terms} = \text{upper limit} - \text{lower limit} + 1\). This becomes \(n\).

Apply the correct sum formula

Use the arithmetic series formula, geometric series formula, or sum to infinity formula if the series converges.

Worked example: Arithmetic sigma (Level 3)

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

1The first term is \(4(1)-19 = -15\), and the common difference is \(4\).
2Use the arithmetic-series sum formula: \(\dfrac{m}{2}[2(-15) + (m-1)4] = 1189\).
3Simplify: \(m(2m-17) = 1189\), so \(2m^2 - 17m - 1189 = 0\).
4Factorise: \((2m+49)(m-29) = 0\), giving \(\mathbf{m=29}\).

Source: 2026 G12 Limpopo CT March, Q2.3.2

Worked example: Geometric sigma with unknown upper limit (Level 3)

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

1The inner series is geometric with \(a = 1\) and \(r = 3\).
2\(\sum_{k=1}^{n}3^{k-1} = \dfrac{3^n - 1}{2}\).
3Substitute: \(\dfrac{5}{3}\cdot \dfrac{3^n - 1}{2} = \dfrac{1820}{3}\).
4This gives \(5(3^n - 1) = 3640 \Rightarrow 3^n = 729 = 3^6\).
5So \(\mathbf{n = 6}\).

Source: 2024 G12 SACAI P1 May_June, Q2.2

Worked example: Infinite sum from sigma (Level 3)

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

1\(T_1 = 8(4)^0 = 8\), \(T_2 = 8(4)^{-1} = 2\), so \(r = \tfrac{2}{8} = \tfrac{1}{4}\).
2Since \(\left|\tfrac{1}{4}\right| < 1\), the series converges.
3\(S_\infty = \dfrac{8}{1-\frac{1}{4}} = \dfrac{8}{\frac{3}{4}} = \mathbf{\dfrac{32}{3}}\)

Worked example: Writing a series in sigma notation

\(1 + 5 + 9 + \cdots + 21\), so \(a = 1\), \(d = 4\), and \(T_k = 4k - 3\).

Find the number of terms first: \(4n - 3 = 21 \Rightarrow n = 6\).

\[\sum_{k=1}^{6}(4k-3)\]

Quick practice: Sigma notation

1. Evaluate \(\displaystyle\sum_{k=1}^{30}(8-5k)\)

2. Find \(m\) if \(\displaystyle\sum_{k=1}^{m}3(2)^{k-1} = 93\)

3. For which values of \(x\) will \(\displaystyle\sum_{k=1}^{\infty}(4x-1)^k\) exist?

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Exam Tips and Common Mistakes

NSC examiner insights. Cognitive levels. Complete formula reference.

Five-step strategy for every sequences question

Use this method on every question

1Identify the type. Check differences or ratio and write your conclusion clearly.
2State the known values. Write \(a = \ldots\), \(d = \ldots\), or \(r = \ldots\), as well as \(n = \ldots\), before substituting.
3Write the full formula. This protects your method marks.
4Substitute carefully. Watch brackets around \((n-1)\) and any negative values.
5Verify your result. If a negative root appears for \(n\), reject it by stating that \(n\) must be a positive natural number.

Seven common mistakes

Incorrect method: Writing \(a + nd\) instead of \(a + (n-1)d\).

Correct method: Write the full formula before substituting.

This is the classic off-by-one error. The first term contains zero copies of the common difference.

Incorrect method: Applying \(S_\infty\) without stating the convergence check.

Correct method: Write \(|r| < 1\) before using the sum to infinity formula.

The convergence check is a method mark in many Paper 1 questions.

Incorrect method: Mixing up arithmetic and geometric formulas.

Correct method: Identify the type first, then choose the formula.

An arithmetic sequence uses a constant difference. A geometric sequence uses a constant ratio.

Incorrect method: Stopping at the first differences for a quadratic sequence.

Correct method: If the first differences are not constant, calculate the second differences.

You need at least 4 terms to confirm a quadratic sequence.

Incorrect method: Miscounting terms in sigma notation.

Correct method: Use upper limit minus lower limit plus 1.

That final \(+1\) is what learners often miss.

Incorrect method: Not checking the term rule after finding it.

Correct method: Substitute \(n = 1\), \(n = 2\), and \(n = 3\) back into the formula.

A quick substitution check can save the rest of a multi-part question.

Incorrect method: Hiding the rejected root when solving for \(n\).

Correct method: Show both roots, then reject the negative root with a reason.

Write: "Since \(n\) must be a positive natural number, reject the negative root."

Cognitive levels: What the exam tests

Level 1: Knowledge

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

Level 2: Routine procedures

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

Level 3: Complex procedures

Handle multi-step questions, linked conditions, and convergence questions with a variable.

Level 4: Problem solving

Work in unfamiliar contexts, combine ideas, and interpret patterns in practical situations.

Complete formula reference

Type	Identify by	General term \(T_n\)	Sum	Notes
Arithmetic sequence and arithmetic series	First difference \(d\) is constant	\(a + (n-1)d\)	\(\dfrac{n}{2}[2a + (n-1)d]\)	You can also use \(\frac{n}{2}(a+l)\) when the last term is known. Also remember \(T_n = S_n - S_{n-1}\).
Quadratic sequence	Second differences are constant	\(an^2 + bn + c\)	No standard formula	\(a = \frac{\text{second difference}}{2}\), \(3a+b = T_2-T_1\), \(a+b+c = T_1\)
Geometric sequence and geometric series	Ratio \(r\) is constant	\(ar^{n-1}\)	\(\dfrac{a(r^n-1)}{r-1}\)	\(\dfrac{a(1-r^n)}{1-r}\) is the equivalent form often used when \(0 < r < 1\).
Geometric series to infinity	\(\|r\| < 1\)	Not required	\(\dfrac{a}{1-r}\)	State the convergence condition before applying the formula.

CAPS exam weighting

Patterns and Sequences usually contributes 25 +/- 3 marks in Paper 1 out of 150 marks. That is roughly 17% of the paper. Both sum formula proofs are directly examinable, and convergence with a variable appears regularly at Level 3 and Level 4.

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Frequently Asked Questions

Short answers to common Grade 12 Patterns, Sequences and Series questions.

FAQ

How do I identify the type of sequence?

Subtract consecutive terms first. If the first differences are constant, it is an arithmetic sequence. If not, subtract again. If the second differences are constant, it is a quadratic sequence. If dividing consecutive terms gives a constant ratio, it is a geometric sequence.

FAQ

When does a geometric series converge?

A geometric series converges only when the absolute value of the common ratio is less than 1. In symbols, \(|r| < 1\).

FAQ

How do I count terms in sigma notation?

Use the rule: number of terms = upper limit - lower limit + 1. For example, \(\sum_{k=3}^{10}\) contains \(10 - 3 + 1 = 8\) terms.

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Patterns, Sequences & SeriesSummary Notes

Arithmetic Sequence and Arithmetic Series

Recognising the three types

Definition and common difference

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

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

Quadratic Sequence

Definition and difference tables

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

Geometric Sequence and Geometric Series

Definition and common ratio

General term and finite sum

Convergence and sum to infinity

Convergent series

Divergent series

Sigma Notation

Anatomy of a sigma expression

Three-step method for every sigma question

Exam Tips and Common Mistakes

Five-step strategy for every sequences question

Seven common mistakes

Cognitive levels: What the exam tests

Level 1: Knowledge

Level 2: Routine procedures

Level 3: Complex procedures

Level 4: Problem solving

Complete formula reference

Frequently Asked Questions

Patterns, Sequences & Series
Summary Notes