17 questions arranged by DBE cognitive level — quadratic number patterns, constant second differences, and finding \(T_n=an^2+bn+c\). Work each one on paper first, then reveal the memo.
17
practice questions
4
cognitive levels
2
real exam citations
100%
independently verified
How to use this bank.
Start at Level 1 and move up — don't jump to Level 4 first.
Always build a difference table first: first differences, then second differences. Don't guess the pattern type.
Remember \(2a\) equals the second difference — not \(a\) directly.
Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved from scratch before publication. Two questions (Q12, Q16) are sourced from the real DBE/provincial exam archive and were independently re-derived and confirmed to match the official memo before being included; every other question is an original, labelled "Equation Station SA Practice Question," per this site's citation policy of never attaching a specific exam attribution unless it has been confirmed against the archived paper.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
20%
Level 1 | Knowledge
Identify and Recall
Classify a pattern, and recall the basic relationships.
Q1Equation Station SA Practice Question2 marks
Classification
Linear or Quadratic?
Determine whether \(5;\;11;\;19;\;29;\;\ldots\) is a linear or a quadratic pattern.
Memo
✓ First differences: \(6;8;10\) — not constant✓ Second differences: \(2;2\) — constant✓ The pattern is quadratic
Q2Equation Station SA Practice Question2 marks
Classification
Not Every Pattern Is Quadratic
Determine whether \(7;\;13;\;19;\;25;\;\ldots\) is a linear or a quadratic pattern.
Memo
✓ First differences: \(6;6;6\) — constant✓ The pattern is linear, not quadratic
Q3Equation Station SA Practice Question2 marks
General term
Finding \(a\) From the Second Difference
A quadratic pattern has a constant second difference of \(8\). Determine the value of \(a\) in \(T_n=an^2+bn+c\).
Memo
✓ \(2a=8\Rightarrow a=4\)
Q4Equation Station SA Practice Question2 marks
General term
Direct Substitution
For \(T_n=3n^2-2n+1\), determine \(T_1\).
Memo
✓ \(T_1=3(1)^2-2(1)+1=3-2+1=2\)
30%
Level 2 | Routine Procedures
Difference Tables and Direct Substitution
Confirm a pattern and find its next term, or evaluate a given \(T_n\).
Q5Equation Station SA Practice Question3 marks
Difference table
Confirm and Extend
Show that \(2;\;9;\;20;\;35;\;\ldots\) is a quadratic pattern, and determine the next term.
Memo
✓ First differences: \(7;11;15\)✓ Second differences: \(4;4\) — constant, so quadratic✓ Next first difference: \(15+4=19\), so next term \(=35+19=54\)
Q6Equation Station SA Practice Question4 marks
General term
Finding \(T_n\)
Determine \(T_n\) for the quadratic pattern \(-2;\;1;\;6;\;13;\;\ldots\)
Memo
✓ First differences: \(3;5;7\); second difference \(2\), so \(a=1\)✓ \(3a+b=T_2-T_1=3\Rightarrow3+b=3\Rightarrow b=0\)✓ \(a+b+c=T_1=-2\Rightarrow1+0+c=-2\Rightarrow c=-3\)✓ \(\boxed{T_n=n^2-3}\)
Q7Equation Station SA Practice Question2 marks
General term
Evaluate a Given Term
For \(T_n=2n^2+n-1\), determine \(T_6\).
Memo
✓ \(T_6=2(36)+6-1=72+6-1=77\)
Q8Equation Station SA Practice Question4 marks
General term
Finding \(T_n\)
Determine \(T_n\) for the quadratic pattern \(6;\;10;\;16;\;24;\;\ldots\)
Memo
✓ First differences: \(4;6;8\); second difference \(2\), so \(a=1\)✓ \(3a+b=4\Rightarrow3+b=4\Rightarrow b=1\)✓ \(a+b+c=6\Rightarrow1+1+c=6\Rightarrow c=4\)✓ \(\boxed{T_n=n^2+n+4}\)
Q9Equation Station SA Practice Question2 marks
General term
Evaluate a Given Term
For \(T_n=n^2-4n+9\), determine \(T_3\).
Memo
✓ \(T_3=9-12+9=6\)
30%
Level 3 | Complex Procedures
Finding \(T_n\) and Solving for \(n\)
Full general-term derivations, and using \(T_n\) to check or find a specific term.
Q10Equation Station SA Practice Question5 marks
General term
Finding \(T_n\) and a Distant Term
A quadratic pattern begins \(3;\;10;\;21;\;36;\;\ldots\) Determine \(T_n\), then use it to find \(T_{10}\).
Memo
✓ First differences: \(7;11;15\); second difference \(4\), so \(a=2\)✓ \(3a+b=7\Rightarrow6+b=7\Rightarrow b=1\)✓ \(a+b+c=3\Rightarrow2+1+c=3\Rightarrow c=0\)✓ \(T_n=2n^2+n\), so \(T_{10}=200+10=\boxed{210}\)
Q11Equation Station SA Practice Question4 marks
Solving for n
Which Term Equals a Given Value?
For \(T_n=n^2+3n-1\), determine which term of the pattern equals \(69\).
Memo
✓ \(n^2+3n-1=69\Rightarrow n^2+3n-70=0\)✓ Factorise: \((n-7)(n+10)=0\), so \(n=7\) or \(n=-10\)✓ Reject \(n=-10\) — term positions are positive. \(\boxed{69\text{ is the 7th term}}\)
Q12KZN, November 20253 marks
General term
Finding Two Unknown Terms
The first four terms of a quadratic number pattern are \(-7;\;m;\;5;\;n\). The constant second difference is \(4\). Prove, by calculation, that \(m=-3\) and \(n=17\).
For \(T_n=3n^2-n-2\), determine \(T_7\), and hence state whether \(138\) is a term of this pattern.
Memo
✓ \(T_7=3(49)-7-2=147-9=138\)✓ Since \(T_7=138\) exactly, \(\boxed{138\text{ is a term}}\) — specifically the 7th term
20%
Level 4 | Problem Solving
Reading From a Diagram, and Extra Unknowns
Combining the method with an extra step: a diagram with no numbers given, or an unknown constant to solve for first.
Q14Equation Station SA Practice Question6 marks
Graph interpretation
Determine \(T_n\) From a Diagram (No Numbers Given)
The diagram below shows the first four patterns in a sequence of dot triangles (no term values are written). Count the dots yourself, determine \(T_n\), then find which pattern number has \(45\) dots.
Memo
✓ Counting dots: \(T_1=1\), \(T_2=3\), \(T_3=6\), \(T_4=10\)✓ First differences \(2;3;4\); second differences \(1;1\) — constant, confirming a quadratic pattern with \(a=\tfrac12\)✓ \(3a+b=T_2-T_1=2\Rightarrow\tfrac32+b=2\Rightarrow b=\tfrac12\); \(a+b+c=T_1=1\Rightarrow\tfrac12+\tfrac12+c=1\Rightarrow c=0\)✓ \(T_n=\dfrac{n(n+1)}{2}\). Set \(=45\): \(n^2+n-90=0\Rightarrow(n-9)(n+10)=0\Rightarrow n=9\)✓ \(\boxed{\text{Pattern 9 has 45 dots}}\)
Q15Equation Station SA Practice Question5 marks
General term
Tiling Pattern, Full Method
The number of hexagonal tiles in successive rows of a display are \(8;\;14;\;22;\;32;\;\ldots\) Determine \(T_n\), then find \(T_{20}\).
Memo
✓ First differences: \(6;8;10\); second difference \(2\), so \(a=1\)✓ \(3a+b=6\Rightarrow3+b=6\Rightarrow b=3\)✓ \(a+b+c=8\Rightarrow1+3+c=8\Rightarrow c=4\)✓ \(T_n=n^2+3n+4\), so \(T_{20}=400+60+4=\boxed{464}\)
Q16Adapted from KZN, November 20256 marks
General term
A Modified Pattern
An original linear pattern is modified so that a new quadratic pattern begins \(2;\;6;\;14;\;26;\;\ldots\) Determine \(T_n\), then find which term of the new pattern equals \(266\).
Memo
✓ First differences: \(4;8;12\); second difference \(4\), so \(a=2\)✓ \(3a+b=4\Rightarrow6+b=4\Rightarrow b=-2\)✓ \(a+b+c=2\Rightarrow2-2+c=2\Rightarrow c=2\)✓ \(T_n=2n^2-2n+2\). Set \(=266\): \(n^2-n-132=0\Rightarrow(n-12)(n+11)=0\Rightarrow \boxed{n=12}\) (reject \(n=-11\))
Q17Equation Station SA Practice Question5 marks
General term
Solving for an Unknown Constant First
The general term of a quadratic pattern is \(T_n=n^2-5n+c\), and \(T_4=16\). Determine the value of \(c\), then find \(T_{10}\).
Memo
✓ \(T_4=16\): \((4)^2-5(4)+c=16\Rightarrow16-20+c=16\Rightarrow c=20\)✓ \(T_n=n^2-5n+20\), so \(T_{10}=100-50+20=\boxed{70}\)