The Fundamental Idea#
A linear pattern grows (or shrinks) at a constant rate. The gap between consecutive terms is always the same — we call this the common difference ($d$).
$$T_1, \quad T_1 + d, \quad T_1 + 2d, \quad T_1 + 3d, \quad \ldots$$If you plot the terms on a graph (term number on the x-axis, term value on the y-axis), you get a straight line. That’s why it’s called “linear.”
1. The General Term Formula#
$$\boxed{T_n = a + (n - 1)d}$$| Symbol | Meaning |
|---|---|
| $a$ | First term ($T_1$) |
| $d$ | Common difference ($T_2 - T_1$) |
| $n$ | Position / term number ($n = 1, 2, 3, \ldots$) |
| $T_n$ | Value of the term at position $n$ |
Why $(n - 1)$?#
To get from the 1st term to the $n$th term, you “jump” $(n-1)$ times (not $n$ times). Think of it like steps: to reach step 5 from step 1, you take 4 steps.
The Expanded Form#
Expanding $T_n = a + (n-1)d$ gives:
$$T_n = a + dn - d = dn + (a - d)$$This form — $T_n = dn + c$ where $c = a - d$ — is often faster in exams.
2. Finding the General Term — Step by Step#
Step 1: Find $d = T_2 - T_1$
Step 2: Identify $a = T_1$
Step 3: Substitute into $T_n = a + (n-1)d$ and simplify
Step 4: Verify by checking at least 2 known terms
Worked Example 1 — Increasing Pattern#
Find the general term: $5;\; 8;\; 11;\; 14;\; \ldots$
$d = 8 - 5 = 3$, $a = 5$
$T_n = 5 + (n-1)(3) = 5 + 3n - 3 = 3n + 2$
Check: $T_1 = 3(1) + 2 = 5$ ✓, $T_4 = 3(4) + 2 = 14$ ✓
Worked Example 2 — Decreasing Pattern#
Find the general term: $20;\; 17;\; 14;\; 11;\; \ldots$
$d = 17 - 20 = -3$, $a = 20$
$T_n = 20 + (n-1)(-3) = 20 - 3n + 3 = 23 - 3n$
Check: $T_1 = 23 - 3 = 20$ ✓, $T_3 = 23 - 9 = 14$ ✓
Notice: When $d < 0$, the pattern decreases. The formula still works — $d$ is simply negative.
Worked Example 3 — Fractional Common Difference#
Find the general term: $\frac{1}{2};\; 2;\; \frac{7}{2};\; 5;\; \ldots$
$d = 2 - \frac{1}{2} = \frac{3}{2}$, $a = \frac{1}{2}$
$T_n = \frac{1}{2} + (n-1)\frac{3}{2} = \frac{1}{2} + \frac{3n}{2} - \frac{3}{2} = \frac{3n - 2}{2}$
Check: $T_2 = \frac{6-2}{2} = \frac{4}{2} = 2$ ✓, $T_4 = \frac{12-2}{2} = \frac{10}{2} = 5$ ✓
Worked Example 4 — Negative Terms#
Find the general term: $-7;\; -3;\; 1;\; 5;\; \ldots$
$d = -3 - (-7) = 4$, $a = -7$
$T_n = -7 + (n-1)(4) = -7 + 4n - 4 = 4n - 11$
Check: $T_1 = 4 - 11 = -7$ ✓, $T_3 = 12 - 11 = 1$ ✓
3. Solving for $n$ — “Which Term Equals…?”#
Set $T_n$ equal to the given value and solve for $n$.
Worked Example 5#
Which term of $5;\; 8;\; 11;\; 14;\; \ldots$ equals 50?
$T_n = 3n + 2 = 50$
$3n = 48$
$n = 16$
50 is the 16th term.
Worked Example 6 — Is the Value a Term?#
Is 100 a term of $7;\; 13;\; 19;\; 25;\; \ldots$?
$d = 6$, $a = 7$, so $T_n = 6n + 1$
$100 = 6n + 1$
$n = \frac{99}{6} = 16.5$
Since $n$ must be a positive integer, 100 is NOT a term in this sequence.
Worked Example 7 — First Negative Term#
Find the first negative term of $20;\; 17;\; 14;\; 11;\; \ldots$
$T_n = 23 - 3n < 0$
$23 < 3n$
$n > 7.\overline{6}$
So $n = 8$ is the first integer value that makes $T_n$ negative.
$T_8 = 23 - 24 = -1$ ✓ (and $T_7 = 23 - 21 = 2 > 0$)
The first negative term is $T_8 = -1$.
4. Finding the Pattern Given Two Terms#
When you’re given two terms (not necessarily $T_1$), use the “jump” logic to find $d$, then work backwards to find $a$.
The Jump Formula#
$$d = \frac{T_m - T_k}{m - k}$$The number of jumps between $T_k$ and $T_m$ is $(m - k)$.
Worked Example 8#
$T_3 = 11$ and $T_7 = 23$. Find the general term.
From $T_3$ to $T_7$ is 4 jumps: $\frac{23 - 11}{7 - 3} = \frac{12}{4} = 3$
$d = 3$
Find $a$: $T_3 = a + 2d \Rightarrow 11 = a + 6 \Rightarrow a = 5$
$T_n = 5 + (n-1)(3) = 3n + 2$
Worked Example 9#
$T_5 = 2$ and $T_{12} = -19$. Find the general term.
$d = \frac{-19 - 2}{12 - 5} = \frac{-21}{7} = -3$
$T_5 = a + 4d$: $2 = a + 4(-3) = a - 12 \Rightarrow a = 14$
$T_n = 14 + (n-1)(-3) = 14 - 3n + 3 = 17 - 3n$
Check: $T_5 = 17 - 15 = 2$ ✓, $T_{12} = 17 - 36 = -19$ ✓
5. Word Problems#
Linear patterns appear in real-world contexts. The key is to identify $a$ (the starting value) and $d$ (the constant change).
Worked Example 10#
A taxi charges R15 flag-fall plus R8 per kilometre. Write a formula for the cost after $n$ kilometres.
$a = T_1 = 15 + 8(1) = 23$ (cost for 1 km)
Actually, let’s model this more carefully:
Cost for $n$ km: $C_n = 15 + 8n$
This IS a linear pattern with $d = 8$ (each extra km costs R8).
$C_1 = 23$, $C_2 = 31$, $C_3 = 39$, … → common difference = 8 ✓
How many kilometres can you travel for R111?
$15 + 8n = 111$
$8n = 96$
$n = 12$ km
Worked Example 11#
A stack of chairs has a height of 80 cm for 1 chair. Each additional chair adds 12 cm. Write a formula for the height of a stack of $n$ chairs and find how many chairs make a stack of 200 cm.
$T_1 = 80$, $d = 12$
$T_n = 80 + (n-1)(12) = 80 + 12n - 12 = 12n + 68$
For 200 cm: $12n + 68 = 200 \Rightarrow 12n = 132 \Rightarrow n = 11$ chairs
6. The Connection to Straight-Line Graphs#
A linear pattern $T_n = dn + c$ has the same form as the straight line equation $y = mx + c$:
| Linear pattern | Straight line |
|---|---|
| $T_n = dn + (a - d)$ | $y = mx + c$ |
| Common difference $d$ | Gradient $m$ |
| “Zeroth term” $(a - d)$ | y-intercept $c$ |
| Term number $n$ | x-value |
| Term value $T_n$ | y-value |
This means:
- If $d > 0$, the pattern increases → the line slopes upward
- If $d < 0$, the pattern decreases → the line slopes downward
- The gradient of the line IS the common difference
Key difference: In a linear pattern, $n$ must be a positive integer (you can only have the 1st term, 2nd term, etc.). On a straight-line graph, $x$ can be any real number. So a linear pattern gives you discrete dots on the line, not the full continuous line.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Confusing $n$ and $T_n$ | $n$ = position (seat number), $T_n$ = value (person in that seat) | $n$ is always a positive integer |
| Getting $d$ wrong | $d = T_2 - T_1$ (later minus earlier). If the pattern decreases, $d$ is negative | Double-check the sign of $d$ |
| Off-by-one: using $nd$ instead of $(n-1)d$ | $T_n = a + (n-1)d$, NOT $a + nd$ | The $(n-1)$ accounts for the first term already being at position 1 |
| Not checking the answer | A single arithmetic error in $a$ or $d$ gives the wrong formula | Always verify with at least 2 known terms |
| Accepting non-integer $n$ | If “which term equals 100?” gives $n = 16.5$, the value is NOT a term | $n$ must be a positive integer for it to be a valid term |
| Forgetting the first negative term needs $n > ...$ | Students solve $T_n < 0$ but forget to round UP to the next integer | Always round up and verify both $T_n$ and $T_{n-1}$ |
💡 Pro Tips for Exams#
1. The Quick Formula#
Instead of expanding every time, use: $T_n = dn + (a - d)$.
For $a = 5$, $d = 3$: $T_n = 3n + (5-3) = 3n + 2$. One line, done.
2. The “Is it a Term?” Test#
Set $T_n = \text{value}$ and solve for $n$. If $n$ is a positive integer → yes, it’s a term. If not → no.
3. Finding Two Unknown Terms#
If given $T_k$ and $T_m$:
- $d = \frac{T_m - T_k}{m - k}$ (difference ÷ number of jumps)
- Back-substitute to find $a$
4. First Negative / First Positive Term#
Solve the inequality $T_n < 0$ (or $T_n > 0$) and round to the next integer in the correct direction.
🔗 Related Grade 10 topics:
- Sketching Graphs — the general term $T_n = dn + c$ is a linear function with gradient $d$
- Solving Equations — finding which term equals a value means solving a linear equation
📌 Where this leads in Grade 11:
- Quadratic Patterns — when the first differences are NOT constant, but the second differences are
- Grade 12 Sequences & Series — arithmetic and geometric sequences, sigma notation, and convergence