The Logic of Constant Growth#
An Arithmetic Sequence is a pattern that grows or shrinks at the same speed every time. We call that constant step the Common Difference ($d$).
The rule is simple: every term minus the one before it gives $d$.
$$d = T_2 - T_1 = T_3 - T_2 = T_4 - T_3 = \dots = T_{n} - T_{n-1}$$Let’s see this with a concrete sequence — $3;\;7;\;11;\;15;\;19;\;\dots$:
| Calculation | Result |
|---|---|
| $T_2 - T_1 = 7 - 3$ | $4$ |
| $T_3 - T_2 = 11 - 7$ | $4$ |
| $T_4 - T_3 = 15 - 11$ | $4$ |
| $T_5 - T_4 = 19 - 15$ | $4$ |
Every single gap is $4$. That’s what makes this arithmetic. The general pattern is always:
$$d = T_n - T_{n-1} \quad \text{(any term minus the previous term)}$$If any of those gaps give a different answer, the sequence is not arithmetic.
| Symbol | Meaning | Example ($3;\;7;\;11;\;15;\;\dots$) |
|---|---|---|
| $a$ | First term ($T_1$) | $3$ |
| $d$ | Common difference | $7 - 3 = 4$ |
| $n$ | Position number (always a positive integer) | $n = 1, 2, 3, \dots$ |
| $T_n$ | Value of the term at position $n$ | $T_4 = 15$ |
Key insight: $d$ can be negative (decreasing pattern), positive (increasing), or zero (constant pattern). Always compute $d = T_2 - T_1$ and keep the sign.
- $20;\;14;\;8;\;2;\;\dots$ → $d = 14 - 20 = -6$ (decreasing)
- $5;\;5;\;5;\;5;\;\dots$ → $d = 5 - 5 = 0$ (constant)
- $3;\;7;\;11;\;15;\;\dots$ → $d = 7 - 3 = 4$ (increasing)
1. The General Term: $T_n = a + (n-1)d$#
Don’t memorise blindly — understand the logic:
| Building block | Why? |
|---|---|
| Start at $a$ | That’s your first term |
| Add $d$ a total of $(n-1)$ times | To reach term $n$ you take $(n-1)$ steps from term 1 |
The Linear Connection#
An arithmetic sequence is really a straight-line function ($y = mx + c$) — but with one critical restriction.
| Sequence language | Straight-line language |
|---|---|
| $d$ (common difference) | $m$ (gradient) |
| $a - d$ (“Term 0”) | $c$ (y-intercept) |
| $n$ (position) | $x$ |
| $T_n$ (value) | $y$ |
This means you can read $d$ off a graph as the gradient, and plotting $(n;\;T_n)$ always gives a straight line.
Discrete vs Continuous — Why $n$ Must Be an Integer#
A straight-line function like $y = 4x + 1$ is continuous — you can plug in $x = 2.7$ or $x = \pi$ and get a valid answer. The line has no gaps.
A sequence is discrete — it only exists at specific, separated points. You can have $T_1$, $T_2$, $T_3$, but there is no “$T_{2.5}$” or “$T_{\frac{1}{3}}$”. The position $n$ must be a natural number ($n \in \mathbb{N}$, i.e. $n = 1, 2, 3, \dots$).
| Feature | Straight line ($y = mx + c$) | Arithmetic sequence ($T_n = a + (n-1)d$) |
|---|---|---|
| Input values | Any real number ($x \in \mathbb{R}$) | Only positive integers ($n \in \mathbb{N}$) |
| Graph | Solid, unbroken line | Separate dots, no line joining them |
| Between points | Infinite values exist | Nothing exists — no $T_{1.5}$ |
Why this matters in exams: If you solve for $n$ and get $n = 7.3$, that’s not a valid term — there is no “term 7.3” in a sequence. Either recheck your work, or the question is asking for the nearest valid term.
For a refresher on the number system ($\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$), see Integers & Number Sense.
Worked Example 1 — Finding the General Term#
$$a = 5,\quad d = 9 - 5 = 4$$$$T_n = a + (n-1)d = 5 + (n-1)(4) = 5 + 4n - 4$$$$\boxed{T_n = 4n + 1}$$Given $5;\;9;\;13;\;17;\;\dots$, determine $T_n$.
Check: $T_1 = 4(1)+1 = 5\;\checkmark\quad T_4 = 4(4)+1 = 17\;\checkmark$
Worked Example 2 — Solving for $n$#
$$T_n = 4n + 1$$$$101 = 4n + 1$$$$4n = 100$$$$\boxed{n = 25}$$Which term of $5;\;9;\;13;\;17;\;\dots$ equals $101$?
$T_{25} = 101$. Since $n$ is a positive integer, this is valid.
Worked Example 3 — Two Terms Given, Find $a$ and $d$#
In an arithmetic sequence, $T_5 = 23$ and $T_{12} = 58$. Find $a$ and $d$.
Step 1 — Use the “Jump Logic”:
The idea: in an arithmetic sequence, every step from one term to the next adds $d$. So if you jump multiple positions, you add $d$ that many times.
From position 5 to position 12 is $12 - 5 = 7$ positions apart, which means 7 common differences separate them:
$$T_{12} = T_5 + 7d$$Rearranging:
$$7d = T_{12} - T_5 = 58 - 23 = 35$$$$\boxed{d = 5}$$Let’s see this jump logic in action across different pairs using the same sequence ($a = 3$, $d = 5$):
| From | To | Positions apart | Calculation | Result |
|---|---|---|---|---|
| $T_1 = 3$ | $T_5 = 23$ | $5 - 1 = 4$ jumps | $4d = 23 - 3 = 20$ | $d = 5\;\checkmark$ |
| $T_3 = 13$ | $T_8 = 38$ | $8 - 3 = 5$ jumps | $5d = 38 - 13 = 25$ | $d = 5\;\checkmark$ |
| $T_5 = 23$ | $T_{12} = 58$ | $12 - 5 = 7$ jumps | $7d = 58 - 23 = 35$ | $d = 5\;\checkmark$ |
| $T_1 = 3$ | $T_{12} = 58$ | $12 - 1 = 11$ jumps | $11d = 58 - 3 = 55$ | $d = 5\;\checkmark$ |
The general formula: $(k - m)$ jumps separate $T_m$ and $T_k$, so $(k-m)d = T_k - T_m$.
Step 2 — Substitute back:
$$T_5 = a + 4d$$$$23 = a + 4(5) = a + 20$$$$\boxed{a = 3}$$Check: $T_{12} = 3 + 11(5) = 58\;\checkmark$
Worked Example 4 — Three Consecutive Terms with Algebra#
The first three terms of an arithmetic sequence are $x + 2;\;2x + 1;\;4x - 3$. Determine $x$ and the sequence.
For an arithmetic sequence the common difference is constant, so:
$$T_2 - T_1 = T_3 - T_2$$$$(2x+1) - (x+2) = (4x-3) - (2x+1)$$$$x - 1 = 2x - 4$$$$-1 + 4 = 2x - x$$$$\boxed{x = 3}$$Substituting: $T_1 = 5,\;T_2 = 7,\;T_3 = 9 \quad\Rightarrow\quad d = 2\;\checkmark$
2. Sequence vs Series — Know the Difference#
Before we calculate sums, you must be crystal clear on the difference between a sequence and a series. They are related but not the same thing.
A sequence is a list of numbers in order. A series is what you get when you add those numbers together.
| Sequence | Series | |
|---|---|---|
| What it is | A list of terms, separated by semicolons | The sum of those terms, joined by $+$ signs |
| Notation | $T_1;\;T_2;\;T_3;\;\dots;\;T_n$ | $T_1 + T_2 + T_3 + \dots + T_n$ |
| Symbol | $T_n$ gives you a single term | $S_n$ gives you the total of $n$ terms |
| Example | $3;\;7;\;11;\;15$ | $3 + 7 + 11 + 15 = 36$ |
| What it tells you | The value at each position | The accumulated total up to that position |
Look at the sequence $2;\;5;\;8;\;11;\;14;\;\dots$ ($a = 2$, $d = 3$):
| $n$ | Term ($T_n$) | Series / Partial sum ($S_n$) | What $S_n$ includes |
|---|---|---|---|
| $1$ | $T_1 = 2$ | $S_1 = 2$ | Just $T_1$ |
| $2$ | $T_2 = 5$ | $S_2 = 2 + 5 = 7$ | $T_1 + T_2$ |
| $3$ | $T_3 = 8$ | $S_3 = 2 + 5 + 8 = 15$ | $T_1 + T_2 + T_3$ |
| $4$ | $T_4 = 11$ | $S_4 = 2 + 5 + 8 + 11 = 26$ | $T_1 + T_2 + T_3 + T_4$ |
| $5$ | $T_5 = 14$ | $S_5 = 2 + 5 + 8 + 11 + 14 = 40$ | $T_1 + \dots + T_5$ |
Notice: each $S_n$ is the running total — it grows every time you add the next term.
Exam language: When a question says “the sequence $3;\;7;\;11;\;\dots$” — they’re talking about the list. When it says “the series $3 + 7 + 11 + \dots$” — they want a sum. Watch for semicolons (sequence) vs plus signs (series) in the question.
3. Arithmetic Series — The Sum: $S_n$#
Now that you understand the difference, let’s find a formula for the sum.
$$1 + 2 + 3 + \dots + 100 = \;?$$Two Forms of the Sum Formula#
| Formula | When to use |
|---|---|
| $S_n = \dfrac{n}{2}\bigl[2a + (n-1)d\bigr]$ | When you know $a$, $d$, and $n$ |
| $S_n = \dfrac{n}{2}(a + l)$ | When you know the first term $a$ and last term $l$ |
Conceptual meaning: $S_n = n \times \text{(average of first and last term)}$. You’re multiplying the number of terms by their average value.
Proof of the Arithmetic Sum Formula (CAPS Required — Full Marking Guideline Version)#
This proof is regularly examined. Marks are awarded for every line, so you must reproduce it exactly. Here is the full version as it appears in textbooks and the official marking guidelines.
Step 1 — Write $S_n$ in full, forwards:
$$S_n = a + (a+d) + (a+2d) + \dots + [a + (n-1)d] \quad \cdots (1)$$Step 2 — Write $S_n$ in reverse order:
The last term is $l = a + (n-1)d$, so writing backwards:
$$S_n = [a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d] + \dots + a \quad \cdots (2)$$Step 3 — Add equation (1) and equation (2), column by column:
Each pair adds to the same value:
| Column | From (1) | From (2) | Sum |
|---|---|---|---|
| 1st | $a$ | $a + (n-1)d$ | $2a + (n-1)d$ |
| 2nd | $a + d$ | $a + (n-2)d$ | $2a + (n-1)d$ |
| 3rd | $a + 2d$ | $a + (n-3)d$ | $2a + (n-1)d$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
| $n$-th | $a + (n-1)d$ | $a$ | $2a + (n-1)d$ |
Every column gives $2a + (n-1)d$, and there are $n$ columns:
$$2S_n = n[2a + (n-1)d]$$Step 4 — Divide both sides by 2:
$$\boxed{S_n = \frac{n}{2}[2a + (n-1)d]}$$Step 5 — Alternative form (substitute $l = a + (n-1)d$):
$$2a + (n-1)d = a + [a + (n-1)d] = a + l$$$$\boxed{S_n = \frac{n}{2}(a + l)}$$Exam tip: Practise writing this proof from memory until it’s automatic. The most common mistake is forgetting to write the series in reverse on a separate line — examiners look for that explicitly.
Worked Example 5 — Basic Sum#
$$a = 3,\quad d = 4,\quad n = 20$$$$S_{20} = \frac{20}{2}\bigl[2(3) + (20-1)(4)\bigr] = 10\bigl[6 + 76\bigr] = 10(82)$$$$\boxed{S_{20} = 820}$$Find the sum of the first 20 terms of $3;\;7;\;11;\;15;\;\dots$
Worked Example 6 — Sum with Last Term Known#
Find: $2 + 5 + 8 + \dots + 302$
Step 1 — Find $n$:
$$T_n = a + (n-1)d \quad\Rightarrow\quad 302 = 2 + (n-1)(3) = 3n - 1$$$$3n = 303 \quad\Rightarrow\quad n = 101$$Step 2 — Use the short formula:
$$S_{101} = \frac{101}{2}(2 + 302) = \frac{101}{2}(304) = 101 \times 152$$$$\boxed{S_{101} = 15\,352}$$Worked Example 7 — Finding $n$ When Given $S_n$#
$$S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$$$$225 = \frac{n}{2}\bigl[2(1) + (n-1)(2)\bigr] = \frac{n}{2}(2 + 2n - 2) = \frac{n}{2}(2n) = n^2$$$$n^2 = 225$$$$\boxed{n = 15} \quad (n = -15 \text{ rejected, } n > 0)$$The sum of an arithmetic series is $S_n = 225$, with $a = 1$ and $d = 2$. How many terms?
Check: The sequence is $1;\;3;\;5;\;7;\;\dots$ (odd numbers). $T_{15} = 1 + 14(2) = 29$.
$$S_{15} = \frac{15}{2}(1 + 29) = \frac{15}{2}(30) = 225\;\checkmark$$Exam warning: If your $n$ turns out to be a decimal, it means no exact whole term achieves that sum. Re-read the question — it may ask for the smallest $n$ such that $S_n >$ some value, in which case round up.
4. The Subtraction Technique: Finding a Term from Sums#
This is one of the most important and most tested techniques. If you are given $S_n$ (as a formula), you can find any individual term:
$$\boxed{T_n = S_n - S_{n-1} \quad\text{for } n \geq 2}$$And separately: $T_1 = S_1$.
Why This Works — See It With Numbers#
Let’s build this up from scratch so you really see it. Take the sequence $2;\;5;\;8;\;11;\;14;\;17;\;20;\;\dots$ ($a = 2$, $d = 3$).
Here are the terms and the running sums side by side:
| $n$ | $T_n$ | $S_n$ (running total) | What $S_n$ actually adds |
|---|---|---|---|
| $1$ | $2$ | $S_1 = 2$ | $2$ |
| $2$ | $5$ | $S_2 = 7$ | $2 + 5$ |
| $3$ | $8$ | $S_3 = 15$ | $2 + 5 + 8$ |
| $4$ | $11$ | $S_4 = 26$ | $2 + 5 + 8 + 11$ |
| $5$ | $14$ | $S_5 = 40$ | $2 + 5 + 8 + 11 + 14$ |
| $6$ | $17$ | $S_6 = 57$ | $2 + 5 + 8 + 11 + 14 + 17$ |
| $7$ | $20$ | $S_7 = 77$ | $2 + 5 + 8 + 11 + 14 + 17 + 20$ |
Now watch what happens when we subtract consecutive sums:
| Subtraction | Calculation | Result | What’s left? |
|---|---|---|---|
| $S_2 - S_1$ | $7 - 2$ | $5$ | That’s $T_2\;\checkmark$ |
| $S_3 - S_2$ | $15 - 7$ | $8$ | That’s $T_3\;\checkmark$ |
| $S_4 - S_3$ | $26 - 15$ | $11$ | That’s $T_4\;\checkmark$ |
| $S_5 - S_4$ | $40 - 26$ | $14$ | That’s $T_5\;\checkmark$ |
| $S_6 - S_5$ | $57 - 40$ | $17$ | That’s $T_6\;\checkmark$ |
| $S_7 - S_6$ | $77 - 57$ | $20$ | That’s $T_7\;\checkmark$ |
Every single time, subtracting the previous sum peels off all the earlier terms and leaves only the term at that position.
The General Logic#
$$S_n = T_1 + T_2 + \dots + T_{n-1} + T_n$$$$S_{n-1} = T_1 + T_2 + \dots + T_{n-1}$$Subtracting: everything cancels except the last term:
$$S_n - S_{n-1} = T_n$$Why $T_1 = S_1$ separately? Because $S_0$ doesn’t exist (there’s no “sum of zero terms” in this context). So for the first term, you just compute $S_1$ directly.
Worked Example 8 — Finding $T_n$ from $S_n$#
Given $S_n = 3n^2 + 5n$, find $T_n$ and show it is arithmetic.
Step 1 — Find $T_1$:
$$T_1 = S_1 = 3(1)^2 + 5(1) = 8$$Step 2 — Find $T_n$ for $n \geq 2$:
$$S_n = 3n^2 + 5n$$$$S_{n-1} = 3(n-1)^2 + 5(n-1)$$$$= 3(n^2 - 2n + 1) + 5n - 5$$$$= 3n^2 - 6n + 3 + 5n - 5$$$$= 3n^2 - n - 2$$$$T_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2$$Step 3 — Verify $T_1$:
$T_1 = 6(1) + 2 = 8 = S_1\;\checkmark$ — so the formula $T_n = 6n + 2$ works for all $n \geq 1$.
Step 4 — Confirm arithmetic:
$d = T_{n+1} - T_n = 6(n+1) + 2 - (6n + 2) = 6$ — constant, so the sequence is arithmetic with $d = 6$.
Worked Example 9 — Using Subtraction to Find a Specific Term#
Given $S_n = n^2 + 4n$, find $T_{10}$.
Method 1 — Direct subtraction:
$$T_{10} = S_{10} - S_9 = \bigl[100 + 40\bigr] - \bigl[81 + 36\bigr] = 140 - 117 = \boxed{23}$$Method 2 — Find general $T_n$ first:
$$T_n = S_n - S_{n-1} = (n^2 + 4n) - ((n-1)^2 + 4(n-1))$$$$= n^2 + 4n - n^2 + 2n - 1 - 4n + 4 = 2n + 3$$$$T_{10} = 2(10) + 3 = \boxed{23}\;\checkmark$$Worked Example 10 — Exam-Style: Given Partial Sums#
In an arithmetic series, $S_5 = 50$ and $S_{10} = 200$. Determine $a$ and $d$.
Step 1 — Set up two equations:
$$S_5 = \frac{5}{2}[2a + 4d] = 5a + 10d = 50 \quad\Rightarrow\quad a + 2d = 10 \quad\cdots(1)$$$$S_{10} = \frac{10}{2}[2a + 9d] = 10a + 45d = 200 \quad\Rightarrow\quad 2a + 9d = 40 \quad\cdots(2)$$Step 2 — Solve simultaneously:
$(2) - 2\times(1)$:
$$2a + 9d - 2a - 4d = 40 - 20$$$$5d = 20 \quad\Rightarrow\quad \boxed{d = 4}$$From $(1)$: $a + 8 = 10 \;\Rightarrow\; \boxed{a = 2}$
Check: $S_5 = 5(2) + 10(4) = 10 + 40 = 50\;\checkmark$ and $S_{10} = 10(2) + 45(4) = 20 + 180 = 200\;\checkmark$
Worked Example 11 — Finding the Sum Between Two Positions#
For the arithmetic sequence with $a = 2$ and $d = 4$, find the sum from the 8th term to the 15th term inclusive.
First, let’s understand exactly what each sum contains. Here are the first 15 terms of this sequence:
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $T_n$ | 2 | 6 | 10 | 14 | 18 | 22 | 26 | 30 | 34 | 38 | 42 | 46 | 50 | 54 | 58 |
Now look at what each sum covers:
| Sum | Terms included | What it adds up |
|---|---|---|
| $S_{15}$ | $T_1$ through $T_{15}$ | $2 + 6 + 10 + \dots + 54 + 58$ |
| $S_7$ | $T_1$ through $T_7$ | $2 + 6 + 10 + \dots + 22 + 26$ |
| $S_{15} - S_7$ | $T_8$ through $T_{15}$ | $30 + 34 + 38 + 42 + 46 + 50 + 54 + 58$ |
Why $S_7$ and not $S_8$? If you subtracted $S_8$, you’d remove $T_8$ along with all the earlier terms — but you want $T_8$ in your answer. So you subtract $S_7$, which removes only terms 1 through 7, leaving terms 8 through 15.
General rule: Sum from $T_p$ to $T_q$ = $S_q - S_{p-1}$. Always subtract one position before where you want to start.
Now the calculation:
$$S_{15} = \frac{15}{2}[2(2) + 14(4)] = \frac{15}{2}[4 + 56] = \frac{15}{2}(60) = 450$$$$S_7 = \frac{7}{2}[2(2) + 6(4)] = \frac{7}{2}[4 + 24] = \frac{7}{2}(28) = 98$$$$\boxed{S_{15} - S_7 = 450 - 98 = 352}$$Quick check: $30 + 34 + 38 + 42 + 46 + 50 + 54 + 58 = 352\;\checkmark$
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Confusing $n$ and $T_n$ | $n$ = position (seat number), $T_n$ = value (person in seat) | Always label what you’re solving for |
| Getting a decimal for $n$ | $n$ must be a positive integer | Recheck your calculation or the question asks “how many terms” |
| Forgetting the sign of $d$ | $10;\;7;\;4;\;\dots$ has $d = -3$, not $3$ | Always compute $d = T_2 - T_1$ |
| Using $S_n - S_n$ instead of $S_n - S_{n-1}$ | That gives zero, not a term | Remember: subtract the previous partial sum |
| Dropping brackets in $S_{n-1}$ | $3(n-1)^2 \neq 3n^2 - 1$ | Expand $(n-1)^2$ fully: $n^2 - 2n + 1$ |
| Sum between positions: using $S_{15} - S_8$ for terms 8 to 15 | This excludes $T_8$ | Use $S_{15} - S_7$ to include $T_8$ |
🎥 Video Lessons#
Watch these videos for a full visual walkthrough of arithmetic sequences and series:
Sequecnes & Series 1#
Sequecnes & Series 2#
💡 Pro Tips for Exams#
1. Jump Logic Shortcut#
If you’re given any two terms $T_m$ and $T_k$ (where $k > m$), you can find $d$ in one step:
$$d = \frac{T_k - T_m}{k - m}$$Why this works: From position $m$ to position $k$, you take $(k - m)$ steps. Each step adds $d$. So the total difference between the two terms is $(k-m) \times d$. Dividing the total difference by the number of steps gives you $d$.
Think of it like speed: if you travel 35 km in 7 hours at a constant speed, your speed is $\frac{35}{7} = 5$ km/h. Same logic — the “speed” of the sequence is $d$.
Example: $T_4 = 19$ and $T_{10} = 49$. Then $d = \dfrac{49 - 19}{10 - 4} = \dfrac{30}{6} = 5$. Done — no simultaneous equations needed.
2. Quadratic $S_n$ Means Linear $T_n$#
If you’re told $S_n = 3n^2 + 5n$ (a quadratic in $n$), then the sequence must be arithmetic and $T_n$ will be linear (first-degree in $n$).
Why? When you compute $T_n = S_n - S_{n-1}$, the $n^2$ terms partially cancel, leaving a linear expression. Specifically, the coefficient of $n^2$ in $S_n$ equals $\frac{d}{2}$. So if $S_n = 3n^2 + \dots$, then $d = 2 \times 3 = 6$.
3. Always Check $S_1 = T_1$#
Whenever you derive $T_n$ from $S_n$, plug $n = 1$ into your $T_n$ formula and check it matches $S_1$.
Why? The formula $T_n = S_n - S_{n-1}$ only works for $n \geq 2$. If your derived $T_n$ formula also works at $n = 1$, great — one formula covers everything. If it doesn’t match $S_1$, the sequence may not be arithmetic, or you need to state $T_1$ separately.
4. Reading the Question Carefully#
| The question says… | What to use |
|---|---|
| “Sum of the first $n$ terms” | $S_n$ directly |
| “Sum from term $p$ to term $q$” | $S_q - S_{p-1}$ |
| “Which term equals…” | Solve $T_n = \text{value}$ for $n$ |
| “How many terms are needed for the sum to exceed…” | Solve $S_n > \text{value}$, round $n$ up |