From Linear to Quadratic — What Changes?#
In Grade 10, you worked with linear patterns where the first differences are constant. The general term was $T_n = dn + c$ — a straight line.
In Grade 11, the pattern accelerates. The terms don’t grow by the same amount each time — they grow by an increasing (or decreasing) amount. This means the first differences are NOT constant, but the second differences are.
| Pattern type | General term | First differences | Second differences |
|---|---|---|---|
| Linear | $T_n = dn + c$ | Constant ($= d$) | All zero |
| Quadratic | $T_n = an^2 + bn + c$ | Changing (linear) | Constant ($= 2a$) |
The test: If the second differences are constant and non-zero, the pattern is quadratic.
1. Understanding the Difference Table#
The difference table is your most important tool. It organises the sequence, its first differences, and its second differences.
Building a Difference Table#
Sequence: $3;\; 7;\; 13;\; 21;\; 31;\; \dots$
| Position | $T_1$ | $T_2$ | $T_3$ | $T_4$ | $T_5$ |
|---|---|---|---|---|---|
| Terms | 3 | 7 | 13 | 21 | 31 |
| 1st differences | 4 | 6 | 8 | 10 | |
| 2nd differences | 2 | 2 | 2 |
The first differences ($T_2 - T_1$, $T_3 - T_2$, etc.) are: $4;\; 6;\; 8;\; 10$ — they’re not constant (so it’s not linear).
The second differences ($6 - 4$, $8 - 6$, $10 - 8$) are: $2;\; 2;\; 2$ — constant! This confirms the pattern is quadratic.
2. WHY $2a$ = Second Difference#
This is the key insight most textbooks skip. Let’s prove it from the formula $T_n = an^2 + bn + c$.
First differences: The difference between consecutive terms is:
$$T_{n+1} - T_n = a(n+1)^2 + b(n+1) + c - (an^2 + bn + c)$$$$= a(n^2 + 2n + 1) + bn + b + c - an^2 - bn - c$$$$= 2an + a + b$$So the first differences are $2an + (a + b)$ — a linear expression in $n$. This is why the first differences form a linear pattern!
Second differences: The difference between consecutive first differences:
$$(2a(n+1) + a + b) - (2an + a + b) = 2a$$The second difference is always $2a$ — a constant. This is why $2a = d_2$ always works.
Deep insight: A quadratic pattern has linear first differences and constant second differences, just like a quadratic function has a linear derivative and a constant second derivative.
3. The Three Equations — Finding $a$, $b$, $c$#
Once you’ve confirmed the pattern is quadratic ($T_n = an^2 + bn + c$), use these three equations in order:
| Step | Equation | What it finds |
|---|---|---|
| 1 | $2a = d_2$ (second difference) | $a$ |
| 2 | $3a + b = d_1$ (first first difference: $T_2 - T_1$) | $b$ |
| 3 | $a + b + c = T_1$ (the first term) | $c$ |
Where does $3a + b = T_2 - T_1$ come from?#
The first first difference is $T_2 - T_1$:
$T_2 - T_1 = (4a + 2b + c) - (a + b + c) = 3a + b$
So the first first difference always equals $3a + b$.
4. Worked Example 1 — Standard Problem#
Find the general term: $3;\; 7;\; 13;\; 21;\; \dots$
Step 1 — Differences:
1st differences: $4;\; 6;\; 8$
2nd differences: $2;\; 2$ → constant, so quadratic ✓
Step 2 — Find $a$:
$2a = 2 \Rightarrow a = 1$
Step 3 — Find $b$:
$3a + b = T_2 - T_1 = 4$
$3(1) + b = 4 \Rightarrow b = 1$
Step 4 — Find $c$:
$a + b + c = T_1 = 3$
$1 + 1 + c = 3 \Rightarrow c = 1$
Step 5 — General term:
$$\boxed{T_n = n^2 + n + 1}$$Step 6 — Verify with at least 2 known terms:
$T_1 = 1 + 1 + 1 = 3$ ✓
$T_3 = 9 + 3 + 1 = 13$ ✓
$T_5 = 25 + 5 + 1 = 31$ ✓
5. Worked Example 2 — Negative Second Difference#
Find the general term: $5;\; 3;\; -1;\; -7;\; \dots$
1st differences: $-2;\; -4;\; -6$
2nd differences: $-2;\; -2$ → constant ✓
$2a = -2 \Rightarrow a = -1$
$3(-1) + b = -2 \Rightarrow b = 1$
$(-1) + (1) + c = 5 \Rightarrow c = 5$
$$T_n = -n^2 + n + 5$$Check: $T_3 = -9 + 3 + 5 = -1$ ✓, $T_4 = -16 + 4 + 5 = -7$ ✓
Notice: $a < 0$ means the terms eventually decrease without bound. The pattern has a maximum value (like an upside-down parabola).
6. Worked Example 3 — Large Second Difference#
Find the general term: $2;\; 10;\; 24;\; 44;\; \dots$
1st differences: $8;\; 14;\; 20$
2nd differences: $6;\; 6$ → constant ✓
$2a = 6 \Rightarrow a = 3$
$3(3) + b = 8 \Rightarrow b = -1$
$3 + (-1) + c = 2 \Rightarrow c = 0$
$$T_n = 3n^2 - n$$Check: $T_2 = 12 - 2 = 10$ ✓, $T_4 = 48 - 4 = 44$ ✓
7. Solving for $n$ — “Which Term Equals…?”#
Once you have the general term, you can find which term has a particular value by solving a quadratic equation.
Worked Example 4#
Which term of $3;\; 7;\; 13;\; 21;\; \dots$ equals $111$?
We found $T_n = n^2 + n + 1$. Set $T_n = 111$:
$n^2 + n + 1 = 111$
$n^2 + n - 110 = 0$
$(n + 11)(n - 10) = 0$
$n = -11$ (rejected — $n$ must be a positive integer) or $n = 10$
111 is the 10th term.
Worked Example 5#
Is 50 a term in the sequence $2;\; 10;\; 24;\; 44;\; \dots$?
$T_n = 3n^2 - n = 50$
$3n^2 - n - 50 = 0$
Using the quadratic formula: $n = \frac{1 \pm \sqrt{1 + 600}}{6} = \frac{1 \pm \sqrt{601}}{6}$
$\sqrt{601} \approx 24.52$, so $n \approx \frac{25.52}{6} \approx 4.25$
Since $n$ is not a positive integer, 50 is NOT a term in this sequence.
8. Finding Terms Given Conditions (No Sequence Given)#
These are the hardest exam questions. You’re given conditions about specific terms or differences and must find the formula.
Worked Example 6#
In a quadratic pattern, $T_2 = 1$, $T_3 = -2$, and the second difference is $-4$. Find $T_1$.
Step 1 — Find $a$:
$2a = -4 \Rightarrow a = -2$
Step 2 — Set up simultaneous equations using the given terms:
$T_2 = a(2)^2 + b(2) + c$: $-8 + 2b + c = 1 \Rightarrow 2b + c = 9$ … (i)
$T_3 = a(3)^2 + b(3) + c$: $-18 + 3b + c = -2 \Rightarrow 3b + c = 16$ … (ii)
Step 3 — Solve simultaneously:
(ii) − (i): $b = 7$
From (i): $c = 9 - 14 = -5$
Step 4 — Find $T_1$:
$T_1 = a + b + c = -2 + 7 - 5 = 0$
Worked Example 7#
The first three terms of a quadratic pattern are $x;\; x + 2;\; x + 6$. Find $x$ and the general term.
Step 1 — Second difference must be constant:
1st differences: $(x + 2) - x = 2$ and $(x + 6) - (x + 2) = 4$
2nd difference: $4 - 2 = 2$ → This gives us $2a = 2$, so $a = 1$. But we need more terms or info to confirm $x$.
Step 2 — Use the three shortcut equations:
$2a = 2 \Rightarrow a = 1$
$3a + b = 2$ (the first first difference) $\Rightarrow b = -1$
$a + b + c = x \Rightarrow 1 - 1 + c = x \Rightarrow c = x$
So $T_n = n^2 - n + x$.
Check: $T_2 = 4 - 2 + x = x + 2$ ✓ and $T_3 = 9 - 3 + x = x + 6$ ✓
The formula is $T_n = n^2 - n + x$, valid for any $x$. If the question gives a fourth term (e.g., $T_4 = 15$), then $16 - 4 + x = 15 \Rightarrow x = 3$.
9. First Differences as a Linear Pattern#
The first differences of a quadratic pattern form a linear pattern. This fact is often tested:
If $T_n = an^2 + bn + c$, then the first difference between $T_n$ and $T_{n+1}$ is:
$$d_n = T_{n+1} - T_n = 2an + (a + b)$$This is a linear expression in $n$ with gradient $2a$ and constant term $(a + b)$.
Worked Example 8#
The first differences of a quadratic pattern are $5;\; 8;\; 11;\; 14;\; \dots$. The first term is $T_1 = 4$. Find the general term.
The first differences are linear with common difference $3$, so $2a = 3 \Rightarrow a = \frac{3}{2}$.
The first first difference is $5$: $3a + b = 5 \Rightarrow \frac{9}{2} + b = 5 \Rightarrow b = \frac{1}{2}$
$a + b + c = T_1 = 4$: $\frac{3}{2} + \frac{1}{2} + c = 4 \Rightarrow c = 2$
$$T_n = \frac{3}{2}n^2 + \frac{1}{2}n + 2$$Check: $T_1 = \frac{3}{2} + \frac{1}{2} + 2 = 4$ ✓, $T_2 = 6 + 1 + 2 = 9$ ✓ (and $9 - 4 = 5$, matching the first first difference ✓)
10. The Connection to Parabolas#
A quadratic pattern $T_n = an^2 + bn + c$ has the same shape as a parabola $y = ax^2 + bx + c$. The only difference: $n$ must be a positive integer, so you only get discrete dots, not a continuous curve.
| Pattern property | Parabola equivalent |
|---|---|
| $a > 0$: terms eventually increase | Parabola opens upward (has a minimum) |
| $a < 0$: terms eventually decrease | Parabola opens downward (has a maximum) |
| Smallest/largest term value | Turning point of the parabola |
| Second difference = $2a$ | Second derivative = $2a$ (calculus, Grade 12) |
Finding the Minimum/Maximum Term#
The turning point of $y = an^2 + bn + c$ is at $n = -\frac{b}{2a}$.
Since $n$ must be a positive integer, check the terms on either side of $-\frac{b}{2a}$ to find the actual minimum or maximum term.
Worked Example 9#
Find the minimum value of $T_n = n^2 - 8n + 20$.
Turning point at $n = \frac{8}{2} = 4$.
$T_4 = 16 - 32 + 20 = 4$
Check neighbours: $T_3 = 9 - 24 + 20 = 5$ and $T_5 = 25 - 40 + 20 = 5$
Minimum value is 4, occurring at the 4th term.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Using first differences for a quadratic | First differences are constant only for linear patterns | Check: are the second differences constant? |
| Getting $b$ wrong | $3a + b = T_2 - T_1$ (the FIRST first difference) | Don’t use the second or third first difference |
| Not rejecting $n \leq 0$ | Term numbers must be positive integers ($n = 1, 2, 3, \dots$) | Always reject negative or fractional $n$ values |
| Not verifying the formula | An arithmetic error in $a$, $b$, or $c$ ruins everything | Check with at least 2 known terms after finding the formula |
| Confusing “second difference” with “second term” | The second difference is the difference between consecutive first differences | Use a difference table to organise your work |
| Forgetting that $T_n = an^2 + bn + c$ (not $an^2 + b$) | The general term always has three constants | Always solve for all three: $a$, $b$, AND $c$ |
| Assuming $n$ starts at 0 | In CAPS, sequences start at $n = 1$ | $T_1$ is the first term, $T_2$ is the second, etc. |
💡 Pro Tips for Exams#
1. The Three Shortcut Equations#
Memorise these — they work EVERY time:
$$2a = d_2 \qquad 3a + b = d_1 \qquad a + b + c = T_1$$Where $d_2$ is the constant second difference and $d_1 = T_2 - T_1$.
2. The Difference Table#
Always draw a difference table before doing anything. It prevents errors and makes the pattern visible:
Terms: T₁ T₂ T₃ T₄
1st diff: d₁ d₂ d₃
2nd diff: d₂₁ d₂₂3. “Which Term” Questions#
These always reduce to solving a quadratic equation. If the answer isn’t a positive integer, the value is NOT a term in the sequence.
4. Quick Sign Check#
$a > 0$ → the pattern eventually increases (terms get bigger and bigger)
$a < 0$ → the pattern eventually decreases (terms get more and more negative)
🔗 Related Grade 11 topics:
- The Parabola — $T_n = an^2 + bn + c$ IS a parabola with $n$ as the input variable
- Quadratic Equations — “which term equals $k$?” means solving a quadratic equation
📌 Grade 10 foundation: Linear Patterns — constant first differences and $T_n = dn + c$
📌 Grade 12 extension: Quadratic Sequences — deeper problems and connections to series