What even is this?
An arithmetic sequence is a list of numbers where you always add (or subtract) the same amount to get from one term to the next. The amount you add each time is called the common difference (d).
Real-life examples: a training plan that adds 3 km per week, cinema rows with 3 extra seats per row, saving $50 more each month. Anywhere the change is constant, that's arithmetic.
The 10-minute version
Low energy? Do just this and you have still won the day. Zero is the only fail.
The one move: read the first formula box and the first worked example above, then do Practice Question 1. Screenshot the win for Nat and stop there.
Four short ways in. The 🎬 cards are waiting on Nat's videos, and each has a ready-to-read film script tucked underneath. 🖨️ Prefer printable notes? 4 styles here.
Each term is found by adding the same value (d) to the one before it. The "staircase" below shows the sequence 3, 7, 11, 15, 19, each step up is exactly 4.
Subtract any term from the next one:
d = t₂ − t₁ = t₃ − t₂ = ...
e.g. 20, 15, 10, 5, ... has d = −5. The sequence is decreasing.
Rather than listing every term to reach the 50th, there's a formula. It works because every term is just the first term with (n−1) lots of d added on.
🔵 Find tₙ
Sub in t₁, n, and d. Calculate.
e.g. Find t₁₀ when t₁=3, d=4:
t₁₀ = 3 + 9×4 = 39
🟢 Find n
Set tₙ = target value. Solve the equation for n.
e.g. When does tₙ = 63?
3+(n−1)×4=63 → n=16
🟡 Find t₁ or d
Sub in what you know. Solve for the unknown variable.
e.g. t₁₂=47, d=4:
t₁+11×4=47 → t₁=3
Sequence: 12, 15, 18, 21, ... | t₁ = 12, d = 3
How many seats are in row 10?
Use tₙ = t₁ + (n−1)d with n = 10:
t₁₀ = 12 + (10−1) × 3 = 12 + 27 = 39 seats
Which row has exactly 42 seats?
Set tₙ = 42 and solve for n:
12 + (n−1) × 3 = 42
(n−1) × 3 = 30
n − 1 = 10
n = 11 ✓ (check: 12 + 10×3 = 42 ✓)
Tap to reveal the answer. Try it yourself first!
General term: tₙ = t₁ + (n−1)d = 5 + (n−1) × 4 = 5 + 4n − 4 = 4n + 1
Check: t₁ = 4(1)+1 = 5 ✓ t₂ = 4(2)+1 = 9 ✓ t₃ = 4(3)+1 = 13 ✓
For the first negative term: tₙ < 0
50 + (n−1)(−6) < 0
50 − 6n + 6 < 0
56 − 6n < 0
n > 56/6 = 9.33...
So n = 10 is the first negative term.
Check: t₉ = 50 + 8×(−6) = 50−48 = 2 (still positive), t₁₀ = 50 + 9×(−6) = 50−54 = −4 (first negative ✓)
Use t₄ = 22 to find t₁:
t₁ + 3d = 22 → t₁ + 15 = 22 → t₁ = 7
Check: 7, 12, 17, 22 ✓ (t₄=22) t₉ = 7 + 8×5 = 47 ✓
Each additional hour adds $35 → d = 35
Sequence: 115, 150, 185, 220, 255, 290, ...
t₆ = 115 + 5 × 35 = 115 + 175 = $290
This is a classic context question, the arithmetic sequence models the total charge after n hours. The call-out fee sets the first term; the hourly rate is the common difference.