The Logic of Acceleration#
In an arithmetic sequence, the gaps between terms are constant — the sequence grows at a steady speed. A quadratic sequence is different: the gaps themselves are changing. The sequence speeds up or slows down.
Arithmetic vs Quadratic — The Core Difference#
| Arithmetic | Quadratic | |
|---|---|---|
| First differences ($T_2 - T_1$, etc.) | Constant | Changing |
| Second differences | Zero (no change in speed) | Constant (steady acceleration) |
| General term | $T_n = an + b$ (linear) | $T_n = an^2 + bn + c$ (quadratic) |
| Graph of $T_n$ vs $n$ | Straight line | Parabola |
The key test: If the first differences are not constant but the second differences are constant, the sequence is quadratic.
1. Understanding Differences — The Difference Table#
This is the foundation. Before you touch any formula, you must be able to build a difference table. Let’s work through $2;\;5;\;10;\;17;\;26;\;\dots$
Step 1 — Write out the terms:
| $n$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $T_n$ | 2 | 5 | 10 | 17 | 26 |
Step 2 — Calculate the first differences (each term minus the previous):
| Gap | $T_2 - T_1$ | $T_3 - T_2$ | $T_4 - T_3$ | $T_5 - T_4$ |
|---|---|---|---|---|
| First difference | $5 - 2 = 3$ | $10 - 5 = 5$ | $17 - 10 = 7$ | $26 - 17 = 9$ |
The first differences are $3;\;5;\;7;\;9$ — they are not constant, so this is not arithmetic.
Step 3 — Calculate the second differences (differences of the first differences):
| Gap | $5 - 3$ | $7 - 5$ | $9 - 7$ |
|---|---|---|---|
| Second difference | $2$ | $2$ | $2$ |
The second differences are constant at $2$. This confirms the sequence is quadratic.
The Full Difference Table Layout#
Here is how to lay it out neatly (this is how markers expect to see it):
$$\begin{array}{ccccccccc} T_1 & & T_2 & & T_3 & & T_4 & & T_5 \\ 2 & & 5 & & 10 & & 17 & & 26 \\ & 3 & & 5 & & 7 & & 9 & \\ & & 2 & & 2 & & 2 & & \end{array}$$- Row 1: The terms
- Row 2: First differences (between consecutive terms)
- Row 3: Second differences (between consecutive first differences)
Another Example — Decreasing Sequence#
Consider $20;\;15;\;12;\;11;\;12;\;\dots$
| $n$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $T_n$ | 20 | 15 | 12 | 11 | 12 |
| First diff | $-5$ | $-3$ | $-1$ | $1$ | |
| Second diff | $2$ | $2$ |
Second differences are constant at $2$ — this is quadratic. Notice the first differences go from negative to positive, meaning the sequence decreases then increases (like a parabola turning).
2. The General Term: $T_n = an^2 + bn + c$#
Where Do the Three Equations Come From?#
Don’t just memorise “$2a$ = second difference.” Let’s derive it so you understand why.
Start with the general term $T_n = an^2 + bn + c$ and compute the first few terms:
| Term | Substitution | Result |
|---|---|---|
| $T_1$ | $a(1)^2 + b(1) + c$ | $a + b + c$ |
| $T_2$ | $a(2)^2 + b(2) + c$ | $4a + 2b + c$ |
| $T_3$ | $a(3)^2 + b(3) + c$ | $9a + 3b + c$ |
Now compute the first differences:
| First difference | Calculation | Result |
|---|---|---|
| $T_2 - T_1$ | $(4a + 2b + c) - (a + b + c)$ | $3a + b$ |
| $T_3 - T_2$ | $(9a + 3b + c) - (4a + 2b + c)$ | $5a + b$ |
Now compute the second difference:
| Second difference | Calculation | Result |
|---|---|---|
| $(T_3 - T_2) - (T_2 - T_1)$ | $(5a + b) - (3a + b)$ | $2a$ |
This gives us the three key equations:
| Equation | What it means | How to use it |
|---|---|---|
| $2a = \text{second difference}$ | The acceleration determines $a$ | Solve this first — always |
| $3a + b = T_2 - T_1$ | The first gap determines $b$ | Substitute your $a$ and solve for $b$ |
| $a + b + c = T_1$ | The starting value determines $c$ | Substitute $a$ and $b$, solve for $c$ |
Why this order matters: You must solve from bottom to top: $a$ first, then $b$, then $c$. Each equation depends on the values found before it. Trying to find $b$ before $a$ is impossible.
Worked Example 1 — Finding the General Term#
Determine the general term of $2;\;5;\;10;\;17;\;26;\;\dots$
Step 1 — Build the difference table:
First differences: $3;\;5;\;7;\;9$
Second differences: $2;\;2;\;2$ (constant $\checkmark$ — quadratic confirmed)
Step 2 — Find $a$:
$$2a = 2 \quad \Rightarrow \quad \boxed{a = 1}$$Step 3 — Find $b$:
$$3a + b = T_2 - T_1 = 3$$$$3(1) + b = 3$$$$\boxed{b = 0}$$Step 4 — Find $c$:
$$a + b + c = T_1 = 2$$$$1 + 0 + c = 2$$$$\boxed{c = 1}$$General term: $T_n = n^2 + 1$
Check: $T_1 = 1 + 1 = 2\;\checkmark \quad T_3 = 9 + 1 = 10\;\checkmark \quad T_5 = 25 + 1 = 26\;\checkmark$
Worked Example 2 — Negative Second Difference#
Find $T_n$ for the sequence $20;\;15;\;12;\;11;\;12;\;\dots$
Step 1 — Difference table:
First differences: $-5;\;-3;\;-1;\;1$
Second differences: $2;\;2;\;2$ (constant $\checkmark$)
Step 2 — Find $a$, $b$, $c$:
$$2a = 2 \quad \Rightarrow \quad a = 1$$$$3a + b = T_2 - T_1 = -5$$$$3 + b = -5 \quad \Rightarrow \quad b = -8$$$$a + b + c = T_1 = 20$$$$1 - 8 + c = 20 \quad \Rightarrow \quad c = 27$$General term: $T_n = n^2 - 8n + 27$
Check: $T_1 = 1 - 8 + 27 = 20\;\checkmark \quad T_4 = 16 - 32 + 27 = 11\;\checkmark$
Worked Example 3 — Fractional Second Difference#
The first four terms of a quadratic sequence are $1;\;2;\;5;\;10;\;\dots$. Find $T_n$.
Step 1 — Difference table:
First differences: $1;\;3;\;5$
Second differences: $2;\;2$ (constant $\checkmark$)
Step 2 — Find $a$, $b$, $c$:
$$2a = 2 \quad \Rightarrow \quad a = 1$$$$3a + b = T_2 - T_1 = 1$$$$3 + b = 1 \quad \Rightarrow \quad b = -2$$$$a + b + c = T_1 = 1$$$$1 - 2 + c = 1 \quad \Rightarrow \quad c = 2$$General term: $T_n = n^2 - 2n + 2$
Check: $T_3 = 9 - 6 + 2 = 5\;\checkmark \quad T_4 = 16 - 8 + 2 = 10\;\checkmark$
3. Solving for $n$ — “Which Term Equals…?”#
When a question gives you a value and asks which term it is, you set $T_n$ equal to that value and solve the resulting quadratic equation.
Worked Example 4 — Solving for $n$#
$$n^2 + 1 = 82$$$$n^2 = 81$$$$n = \pm 9$$Given $T_n = n^2 + 1$, which term equals $82$?
Since $n$ must be a positive integer: $\boxed{n = 9}$
Check: $T_9 = 81 + 1 = 82\;\checkmark$
Worked Example 5 — Solving for $n$ (Quadratic Formula Needed)#
$$n^2 - 8n + 27 = 12$$$$n^2 - 8n + 15 = 0$$$$(n - 3)(n - 5) = 0$$$$n = 3 \quad \text{or} \quad n = 5$$Given $T_n = n^2 - 8n + 27$, which term equals $12$?
Both are positive integers, so both are valid: $T_3 = 12$ and $T_5 = 12$.
Check: $T_3 = 9 - 24 + 27 = 12\;\checkmark \quad T_5 = 25 - 40 + 27 = 12\;\checkmark$
Why two answers? A quadratic sequence follows a parabola. Just like a parabola can cross a horizontal line at two points, a quadratic sequence can have the same value at two different positions. This is perfectly valid — mention both in your answer.
4. Finding a Specific Term Given Three Consecutive Terms with Algebra#
This is a common exam question: you’re given three consecutive terms in terms of $x$, and you must find $x$.
The Key Principle#
For any three consecutive terms $T_1;\;T_2;\;T_3$ of a quadratic sequence, the second difference is constant. That means:
$$(T_3 - T_2) - (T_2 - T_1) = \text{constant}$$If you only have three terms, you have exactly one second difference, which you set equal to the known second difference (or use the fact that consecutive second differences are equal if you have more terms).
Worked Example 6 — Three Consecutive Terms#
The first three terms of a quadratic sequence are $x;\;5;\;2x + 3$. The second difference is $4$. Find $x$.
Step 1 — Write the first differences:
$$T_2 - T_1 = 5 - x$$$$T_3 - T_2 = (2x + 3) - 5 = 2x - 2$$Step 2 — The second difference equals 4:
$$(T_3 - T_2) - (T_2 - T_1) = 4$$$$(2x - 2) - (5 - x) = 4$$$$2x - 2 - 5 + x = 4$$$$3x - 7 = 4$$$$3x = 11$$$$\boxed{x = \frac{11}{3}}$$Check: Terms are $\frac{11}{3};\;5;\;\frac{31}{3}$. First differences: $\frac{4}{3};\;\frac{16}{3}$. Second difference: $\frac{16}{3} - \frac{4}{3} = 4\;\checkmark$
Worked Example 7 — Four Terms with Unknown#
The first four terms of a quadratic sequence are $3;\;x;\;11;\;y$. The second differences are all equal to $2$. Find $x$ and $y$.
Step 1 — Use the first differences and second differences:
First differences: $(x - 3);\;(11 - x);\;(y - 11)$
Second differences (must all be $2$):
$$(11 - x) - (x - 3) = 2$$$$11 - x - x + 3 = 2$$$$14 - 2x = 2$$$$\boxed{x = 6}$$$$(y - 11) - (11 - x) = 2$$$$(y - 11) - (11 - 6) = 2$$$$y - 11 - 5 = 2$$$$\boxed{y = 18}$$Check: Sequence: $3;\;6;\;11;\;18$. First differences: $3;\;5;\;7$. Second differences: $2;\;2\;\checkmark$
5. The Connection to Parabolas#
A quadratic sequence $T_n = an^2 + bn + c$ has the exact same form as a parabola $y = ax^2 + bx + c$ — but remember, just like arithmetic sequences are discrete (only integer values of $n$), quadratic sequences are also discrete.
| Feature | Parabola ($y = ax^2 + bx + c$) | Quadratic sequence ($T_n = an^2 + bn + c$) |
|---|---|---|
| Input | Any real $x$ | Only $n \in \mathbb{N}$ ($1, 2, 3, \dots$) |
| Graph | Smooth curve | Separate dots |
| $a > 0$ | Opens upward | Sequence eventually increases |
| $a < 0$ | Opens downward | Sequence eventually decreases |
| Turning point | Minimum/maximum of curve | Position where sequence changes direction |
Exam connection: If $a > 0$, the second differences are positive and the sequence eventually grows without bound. If $a < 0$, the second differences are negative and the sequence eventually decreases without bound.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Solving for $b$ before $a$ | The equation $3a + b = T_2 - T_1$ requires knowing $a$ first | Always solve in order: $a$ → $b$ → $c$ |
| Using first difference = $2a$ | The second difference is $2a$, not the first | First difference is $3a + b$ |
| Wrong sign on second difference | $20;\;15;\;12;\;11\;\dots$ has second diff $= +2$, not $-2$ | Always compute differences left to right: later minus earlier |
| Forgetting to check both roots for $n$ | A quadratic equation can give two valid positive integer roots | Both values are valid — the sequence can hit the same value twice |
| Not verifying $T_1$ | If $a + b + c \neq T_1$, you’ve made an error | Always check your formula against the original terms |
🎥 Video Lessons#
Quadratic Sequences 1#
Quadratic Sequences 2#
💡 Pro Tips for Exams#
1. The “Half the Second Difference” Check#
The value of $a$ is always exactly half the second difference: $a = \frac{\text{second diff}}{2}$.
Why? Because we proved that $2a = \text{second difference}$, so dividing by 2 gives $a$. If your second difference is $6$, then $a = 3$. If it’s $-4$, then $a = -2$. Use this to instantly check your $a$ value.
2. Quick Verification Strategy#
After finding $T_n = an^2 + bn + c$, don’t just check $T_1$ — check at least three terms (e.g., $T_1$, $T_3$, and $T_5$). If all match, you’re almost certainly correct. If one doesn’t match, trace back through your $a \to b \to c$ calculations.
3. When the Question Gives You $T_n$ and Asks “Is This Quadratic?”#
Compute $T_1, T_2, T_3, T_4$ from the formula, then build a difference table. If the second differences are constant, confirm it’s quadratic. The second difference will be $2a$ (where $a$ is the coefficient of $n^2$).
4. Counting Terms and Positions#
Remember: $n$ is a position number (a natural number). If solving $T_n = k$ gives $n = 4.5$, there is no term with that value. If the question asks “is $k$ a term in the sequence?”, the answer is no.
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