Quadratic Patterns
Master quadratic number patterns — understand WHY second differences are constant, how to derive the general term, how to find specific terms from given conditions, and how to solve exam-style problems — with full worked examples.
Number Patterns
Table of Contents
Number Patterns: When the Differences Aren’t Constant#
In Grade 10, you worked with linear patterns where the difference between consecutive terms was constant ($d$). In Grade 11, the first differences change — but the second differences are constant. This means the general term is quadratic: $T_n = an^2 + bn + c$.
How to Recognise a Quadratic Pattern#
| Pattern type | First differences | Second differences | General term |
|---|---|---|---|
| Linear (Grade 10) | Constant | N/A | $T_n = dn + c$ |
| Quadratic (Grade 11) | Changing | Constant | $T_n = an^2 + bn + c$ |
Example: The sequence $2;\, 6;\, 14;\, 26;\, \ldots$
| Terms | $T_1 = 2$ | $T_2 = 6$ | $T_3 = 14$ | $T_4 = 26$ |
|---|---|---|---|---|
| First differences | $4$ | $8$ | $12$ | |
| Second differences | $4$ | $4$ ✓ |
Second differences are constant ($= 4$), so this IS a quadratic pattern.
Finding $a$, $b$, and $c$#
The three shortcut equations that work every time:
| Equation | What it uses |
|---|---|
| $2a = d_2$ (second difference) | Gives you $a$ immediately |
| $3a + b = d_1$ (first first-difference) | Use $a$ to find $b$ |
| $a + b + c = T_1$ | Use $a$ and $b$ to find $c$ |
From the example above: $d_2 = 4$, $d_1 = 4$, $T_1 = 2$
- $2a = 4 \Rightarrow a = 2$
- $3(2) + b = 4 \Rightarrow b = -2$
- $2 + (-2) + c = 2 \Rightarrow c = 2$
General term: $T_n = 2n^2 - 2n + 2$
Check: $T_3 = 2(9) - 2(3) + 2 = 18 - 6 + 2 = 14$ ✓
Solving for $n$ (“Which term equals…?”)#
Setting $T_n = k$ gives a quadratic equation. Use the quadratic formula and remember: $n$ must be a positive integer.
Example: Which term of $T_n = 2n^2 - 2n + 2$ equals $82$?
$2n^2 - 2n + 2 = 82$
$2n^2 - 2n - 80 = 0$
$n^2 - n - 40 = 0$
$(n - \frac{1 + \sqrt{161}}{2})$ … Using the formula: $n = \frac{1 \pm \sqrt{1 + 160}}{2} = \frac{1 \pm \sqrt{161}}{2}$
$n \approx \frac{1 + 12.69}{2} \approx 6.84$ — not a whole number, so no term equals exactly 82.
Deep Dive#
- Quadratic Patterns: Full Worked Examples — step-by-step method, finding the general term, solving for $n$, and common exam question types
🚨 Common Mistakes#
- Confusing first and second differences: First differences = gaps between terms. Second differences = gaps between the first differences. You need TWO levels.
- Using $d_1$ incorrectly: $d_1$ is the FIRST first-difference ($T_2 - T_1$), not any random one.
- $n$ must be a positive integer: If solving $T_n = k$ gives $n = 3.5$, then no term equals $k$.
- Not checking the answer: After finding $T_n$, substitute $n = 1, 2, 3$ to verify you get the original sequence.
🔗 Related Grade 11 topics:
- The Parabola — $T_n = an^2 + bn + c$ IS a parabola with $n$ as the input
- Quadratic Equations — “which term equals $k$?” means solving a quadratic
📌 Grade 10 foundation: Linear Patterns
📌 Grade 12 extension: Sequences & Series — arithmetic & geometric sequences, sigma notation, convergence
⏮️ Equations & Inequalities | 🏠 Back to Grade 11 | ⏭️ Functions