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Patterns & Sequences — Grade 11

Learn quadratic number patterns — sequences with a constant second difference — and how to find the general term Tn = an² + bn + c. Notes, past papers and a quiz — everything for this topic is one click away.

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Grade 11 CAPS Mathematics

Patterns & Sequences

Quadratic number patterns: constant second differences, and finding \(T_n=an^2+bn+c\).

What You Already Know From Grade 10

Grade 10 taught you one type of pattern. This lesson adds a second.

Linear (arithmetic) pattern

A sequence where the first differences are constant. That constant is called \(d\), the common difference.

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

\(3;\;7;\;11;\;15;\;\ldots\) has \(d=4\), so \(T_n=3+(n-1)(4)=4n-1\).

What's missing

Not every real exam pattern has a constant first difference. This lesson covers what to do when it doesn't.

Quadratic Number Patterns

When the first differences aren't constant, check the second differences before giving up.

Four triangular dot patterns showing 1, 3, 6 and 10 dots Four small triangular arrangements of dots, with 1, 3, 6 and 10 dots respectively, each row of a triangle having one more dot than the row above it. T₁=1 T₂=3 T₃=6 T₄=10
Reading the diagram: the terms are \(1;\;3;\;6;\;10;\ldots\) — first differences \(2;3;4\) are not constant, but the second differences \(1;1\) are.
Definition

A sequence where the second differences are constant. The first differences themselves form a linear (arithmetic) pattern.

General term
\[T_n=an^2+bn+c\]
Why 4 terms, not 3
Three terms give you only one second difference — you can't yet tell whether it will stay constant. You need at least 4 terms (giving two second differences) before you can confirm a pattern is quadratic.
Term-to-term vs. position-to-term
For a linear pattern, "add \(d\) each time" fully describes it. A quadratic pattern's step size keeps changing (that's the whole point of the second difference) — so a term-to-term description alone can't jump straight to, say, \(T_{50}\). You need the position-to-term formula \(T_n=an^2+bn+c\) for that.
Worked example Level 1

Show that \(5;\;12;\;23;\;38;\;\ldots\) is a quadratic pattern, and find the next term.

Position\(T_1\)\(T_2\)\(T_3\)\(T_4\)
Terms5122338
First differences71115
Second differences44
Show solution
  1. 1First differences: \(12-5=7\), \(23-12=11\), \(38-23=15\). Not constant.
  2. 2Second differences: \(11-7=4\), \(15-11=4\). Constant — confirmed quadratic.
  3. 3Next first difference continues the pattern of adding 4: \(15+4=19\).
  4. 4So \(T_5=38+19=\boxed{57}\).
Quick Check

A quadratic number pattern is identified by:

How many terms do you need, at minimum, to confirm a pattern is quadratic?

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

Three relationships, solved in order: \(a\), then \(b\), then \(c\) — and they're not arbitrary, they fall straight out of the formula.

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

These come from substituting \(n=1,2,3\) into \(T_n=an^2+bn+c\) and taking differences.

Show where these come from
  1. 1Substitute \(n=1,2,3\): \(T_1=a+b+c\), \(T_2=4a+2b+c\), \(T_3=9a+3b+c\).
  2. 2First gap: \(T_2-T_1=(4a+2b+c)-(a+b+c)=3a+b\).
  3. 3Second gap: \(T_3-T_2=(9a+3b+c)-(4a+2b+c)=5a+b\).
  4. 4Second difference \(=(T_3-T_2)-(T_2-T_1)=(5a+b)-(3a+b)=2a\).
  5. 5That's why the second difference is always exactly \(2a\), and why \(3a+b\) is always \(T_2-T_1\) — both come directly from the formula, not a memorised trick.
Common mistake
The second difference equals \(\boldsymbol{2a}\), not \(a\). Forgetting to halve it is the single most common error at this level.
Worked example — a clean \(a=1\) case Level 1

For \(2;\;5;\;10;\;17;\;\ldots\), determine \(T_n\).

Show solution
  1. 1The second difference is \(2\), so \(2a=2 \Rightarrow a=1\).
  2. 2\(T_2-T_1=5-2=3\), so \(3a+b=3 \Rightarrow 3+b=3 \Rightarrow b=0\).
  3. 3\(T_1=2\), so \(a+b+c=2 \Rightarrow 1+0+c=2 \Rightarrow c=1\).
  4. 4So \(\boxed{T_n=n^2+1}\). Check \(n=4\): \(16+1=17\) — matches.
Worked example Level 3

For \(5;\;12;\;23;\;38;\;\ldots\) (from the previous slide), determine \(T_n\).

Show solution
  1. 1The second difference is \(4\), so \(2a=4 \Rightarrow a=2\).
  2. 2\(T_2-T_1=12-5=7\), so \(3a+b=7 \Rightarrow 6+b=7 \Rightarrow b=1\).
  3. 3\(T_1=5\), so \(a+b+c=5 \Rightarrow 2+1+c=5 \Rightarrow c=2\).
  4. 4So \(\boxed{T_n=2n^2+n+2}\). Check \(n=4\): \(2(16)+4+2=38\) — matches.
Quick Check

If the second difference of a quadratic pattern is 10, then \(a\) equals:

A quadratic pattern has \(T_1=4\), \(T_2=9\), and second difference \(2\). What is \(b\)?

Critique a claimed rule Level 1

A learner says the pattern \(4;\;9;\;16;\;25;\;\ldots\) has general term \(T_n=n^2+3\). Is this correct?

Show solution
  1. 1Test \(n=1\): \(1^2+3=4\). Matches \(T_1=4\).
  2. 2Test \(n=2\): \(2^2+3=7\). But the actual \(T_2=9\) — the rule is wrong. It only happened to work for the first term.
  3. 3Find the real rule: second difference is \(2\), so \(a=1\); \(T_2-T_1=5\), so \(3(1)+b=5 \Rightarrow b=2\); \(T_1=4\), so \(1+2+c=4 \Rightarrow c=1\).
  4. 4So \(\boxed{T_n=n^2+2n+1}\). Check \(n=4\): \(16+8+1=25\) ✓
Are these two rules the same? Level 3

Learner A says \(T_n=n^2+2n+1\). Learner B says \(T_n=(n+1)^2\). Are they equivalent?

Show solution
  1. 1Expand Learner B's rule: \((n+1)^2=n^2+2n+1\).
  2. 2That's identical to Learner A's rule — yes, they're the same, just written differently.
Substituting into the general formula directly Level 1

A quadratic pattern has \(a=1\), \(T_1=3\), and \(T_2=6\). Determine \(T_n\).

Show solution
  1. 1Substitute directly into \(T_n=an^2+bn+c\): \(T_1=a+b+c\) and \(T_2=4a+2b+c\).
  2. 2\(T_1=3\): \(1+b+c=3 \Rightarrow b+c=2\)
  3. 3\(T_2=6\): \(4+2b+c=6 \Rightarrow 2b+c=2\)
  4. 4Subtract the first equation from the second: \(b=0\), so \(c=2\).
  5. 5So \(\boxed{T_n=n^2+2}\). Check: \(T_1=1+2=3\) ✓, \(T_2=4+2=6\) ✓
Working backwards, non-adjacent terms Level 3-4

A quadratic pattern has \(a=2\), \(T_1=7\), and \(T_3=29\). Determine \(T_n\).

Show solution
  1. 1This time you're not given \(T_2\) or the second difference directly, so go back to the formula itself: \(T_1=a+b+c\) and \(T_3=9a+3b+c\).
  2. 2\(T_1=7\): \(2+b+c=7 \Rightarrow b+c=5\)
  3. 3\(T_3=29\): \(18+3b+c=29 \Rightarrow 3b+c=11\)
  4. 4Subtract the first equation from the second: \(2b=6 \Rightarrow b=3\), so \(c=5-3=2\).
  5. 5So \(\boxed{T_n=2n^2+3n+2}\). Check: \(T_1=2+3+2=7\) ✓, \(T_3=18+9+2=29\) ✓
Why this matters
The three-relationship shortcut needs \(T_1\), \(T_2\) and the second difference. When you're given different information — like two non-adjacent terms — go back to substituting into \(T_n=an^2+bn+c\) directly and solve simultaneously instead.
Quick Check

A learner claims \(2;\;7;\;14;\;23;\;\ldots\) has \(T_n=n^2+n\). Testing \(n=1\) gives \(2\), which matches \(T_1\). Is the rule confirmed correct?

You're given \(a=3\), \(T_1=10\) and \(T_4=61\) for a quadratic pattern — but not \(T_2\) or the second difference. What should you do?

Worked example — read from a diagram Level 1

The diagram below shows the first four terms of a pattern. No numeric list is given — count the dots yourself, then determine \(T_n\).

Four triangular dot patterns for a diagram-reading question Four triangular arrangements of dots growing from 1 dot to 10 dots, matching a triangular-number pattern. Pattern 1 Pattern 2 Pattern 3 Pattern 4
Show solution
  1. 1Count the dots: \(T_1=1\), \(T_2=3\), \(T_3=6\), \(T_4=10\).
  2. 2First differences: \(2;3;4\). Second differences: \(1;1\) — constant, so \(2a=1 \Rightarrow a=\tfrac{1}{2}\).
  3. 3\(T_2-T_1=2\), so \(3a+b=2 \Rightarrow \tfrac{3}{2}+b=2 \Rightarrow b=\tfrac{1}{2}\).
  4. 4\(T_1=1\), so \(a+b+c=1 \Rightarrow \tfrac{1}{2}+\tfrac{1}{2}+c=1 \Rightarrow c=0\).
  5. 5So \(\boxed{T_n=\tfrac{1}{2}n^2+\tfrac{1}{2}n=\dfrac{n(n+1)}{2}}\) — the well-known triangular-number pattern.

Patterns and Graphs: Points on a Parabola

Plotting position against value turns a quadratic pattern into a curve, not a line.

The points (1,4), (2,9), (3,16) and (4,25) lying on a smooth upward curve Four points plotted with position n on the horizontal axis and term value Tn on the vertical axis, all lying exactly on a smooth upward-curving parabola, for the pattern 4, 9, 16, 25. n Tₙ (1,4) (2,9) (3,16) (4,25)
Reading the diagram: the pattern \(4;9;16;25;\ldots\) has \(T_n=n^2+2n+1\). Plotting \((n,T_n)\) gives points that curve upward — not a straight line. That curve is a parabola, the same shape you'll meet in Functions & Graphs.
Why a curve, not a line?

\(T_n=an^2+bn+c\) has the same structure as a quadratic function \(y=ax^2+bx+c\), with \(n\) playing the role of \(x\). Squaring \(n\) is what bends the line into a parabola.

Steady growth vs. accelerating growth

A straight line means the value grows by the same amount each step (linear). A curve bending upward means the value grows by a larger amount each step — that's what a constant second difference actually looks like.

Worked example — a real-life pattern Level 1

Tickets sold for a school play grow faster each day as word spreads: \(8;\;14;\;22;\;32;\;\ldots\) tickets on days 1 to 4. Determine \(T_n\), then predict day 6's sales.

Show solution
  1. 1First differences: \(6;8;10\). Not constant — sales are accelerating, not growing steadily.
  2. 2Second difference: \(2\), constant — confirmed quadratic. \(2a=2 \Rightarrow a=1\).
  3. 3\(T_2-T_1=6\), so \(3(1)+b=6 \Rightarrow b=3\).
  4. 4\(T_1=8\), so \(1+3+c=8 \Rightarrow c=4\). So \(\boxed{T_n=n^2+3n+4}\).
  5. 5Day 6: \(T_6=36+18+4=\boxed{58\text{ tickets}}\).
Quick Check

Plotting \((n,T_n)\) for a quadratic pattern gives points that lie on:

Using \(T_n=n^2+3n+4\) (the ticket-sales pattern), on which day are exactly 22 tickets sold?

Worked example — direct substitution Level 1

A quadratic pattern has \(T_n=n^2+3\). Determine \(T_5\).

Show solution
  1. 1Substitute \(n=5\): \(T_5=5^2+3=25+3\)
  2. 2\(\boxed{T_5=28}\)
Worked example — find Tn, then solve for n Level 3-4

A quadratic pattern begins \(-3;\;2;\;9;\;18;\;29;\;\ldots\) Determine \(T_n\), then find which term equals \(74\).

Show solution
  1. 1First differences: \(5;7;9;11\). Second difference: \(2\), so \(a=1\).
  2. 2\(T_2-T_1=5\), so \(3(1)+b=5 \Rightarrow b=2\).
  3. 3\(T_1=-3\), so \(1+2+c=-3 \Rightarrow c=-6\). So \(T_n=n^2+2n-6\).
  4. 4Set \(T_n=74\): \(n^2+2n-6=74 \Rightarrow n^2+2n-80=0\).
  5. 5Factorise: \((n-8)(n+10)=0\), so \(n=8\) or \(n=-10\).
  6. 6Reject \(n=-10\) (a term position can't be negative). So \(74\) is the \(\boxed{8\text{th}}\) term.
Quick Check

When solving \(T_n=k\) for \(n\) gives two roots, one positive and one negative, you should:

For \(T_n=n^2-n+1\), what is \(T_5\)?

Putting It Together

Every Grade 11 pattern question follows this exact decision process.

StepWhat to do
1Find the first differences. Constant? → linear pattern, use \(T_n=a+(n-1)d\).
2Not constant? Find the second differences. Constant? → quadratic pattern.
3Given \(T_1\), \(T_2\) and the second difference? Use \(2a=\)second difference, then \(3a+b=T_2-T_1\), then \(a+b+c=T_1\).
4Given different information (e.g. two non-adjacent terms)? Substitute directly into \(T_n=an^2+bn+c\) and solve simultaneously instead.
5To find a specific term, substitute \(n\). To find which term equals a value, solve the quadratic equation and reject any negative or non-integer root.
Looking ahead
Grade 12 builds on this by introducing geometric sequences (constant ratio, not difference) and series — adding up the terms of a sequence, including sigma notation and the sum to infinity.
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What CAPS Expects You to Know

The Grade 11 Patterns knowledge statement this page is built from.

  1. 1

    Investigate number patterns leading to those where there is a constant second difference between consecutive terms, and the general term is therefore quadratic.

How to Use This Lesson

A few practical notes before you start.

  • Make sure your Grade 10 linear-pattern method (\(T_n=a+(n-1)d\)) is solid first — the first differences of every quadratic pattern form a linear pattern.
  • Always build the difference table before writing any formula — it shows you which type of pattern you're dealing with.
  • Remember \(2a\), not \(a\), equals the second difference — the single most common slip at this level.
  • Try each worked example yourself before pressing “Show solution.”

Learn It in Short Videos

Four free, independent videos on the core Grade 11 skill — not made by Equation Station SA.

Why We Halve the Second Difference

The reasoning behind \(2a=\)second difference — understanding this makes the method much easier to remember.

Beast Mode Maths · Why do we half the second difference in a Quadratic Sequence

Finding the General Term

A full worked method for finding \(T_n=an^2+bn+c\) from a given sequence, start to finish.

Radford Mathematics · Quadratic Sequences: how to find the formula for the n-th term

A Full Worked Introduction

A complete walkthrough from spotting a quadratic pattern to writing down its general term.

Kevinmathscience · Quadratic number patterns grade 11: introduction and examples

Another Angle on the Method

A second full explanation of quadratic number patterns — useful if the first video's pace or style didn't click.

Lisa Oswald · Grade 11 Quadratic Number Patterns

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Using \(a\) instead of \(2a\)

Setting \(a\) equal to the second difference directly, instead of halving it first.

Declaring "quadratic" too early

Calling a pattern quadratic after checking only one second difference — always confirm with at least two.

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

Both roots of the quadratic equation are found, but only a positive whole number is a valid term position.

Losing track of positions in the difference table

Writing differences without labelling which \(T_n\) they sit between makes it easy to substitute the wrong values.

Trusting a claimed rule after testing only one term

A wrong rule can still match \(T_1\) by coincidence — always check at least two terms before accepting a rule as correct.

This Topic in Other Grades

Patterns & Sequences builds across all three FET grades — revise or preview the other levels here.

Grade 10

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

Study this

Grade 12

Geometric sequences and series, sigma notation, sum to infinity

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 11 Past Papers

Exam-style Grade 11 quadratic number pattern 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 11 CAPS questions about quadratic number patterns.

What is new in Grade 11 Patterns & Sequences?

Grade 11 introduces quadratic number patterns: sequences whose first differences are not constant, but whose second differences are constant. The general term of such a pattern is quadratic, Tn = an^2 + bn + c.

How many terms do I need to confirm a quadratic pattern?

At least 4 terms. Three terms only give you one second difference, so you cannot yet tell whether it will stay constant.

How do I find a, b and c in Tn = an^2+bn+c?

Use three relationships in order: 2a equals the constant second difference, 3a+b equals the first difference between T2 and T1, and a+b+c equals T1. Solve for a first, then b, then c.

Do I still need Grade 10 linear patterns?

Yes. Grade 10's linear (arithmetic) pattern, Tn = a+(n-1)d, is revised, not retaught, because the first differences of a quadratic pattern always form a linear pattern themselves.

What if I'm not given T1, T2 and the second difference?

Substitute the term positions you do have directly into Tn = an^2+bn+c to get equations in a, b and c, then solve them simultaneously. The three-relationship shortcut is just a faster route to the same answer when T1, T2 and the second difference happen to be given.

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.