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Unit 3 · Sequences & Finance

📐 Arithmetic Sequences

A sequence where the gap between consecutive terms is always the same. One formula, tₙ = a + (n−1)d, lets you find any term, find which term hits a target value, and add up a whole run of terms.

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.

1
Spotting an Arithmetic Sequence

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.

3 t₁ 7 t₂ 11 t₃ 15 t₄ 19 t₅ +4 +4 +4 +4 d = 4
To find d:
Subtract any term from the next one:
d = t₂ − t₁ = t₃ − t₂ = ...
d can be negative!
e.g. 20, 15, 10, 5, ... has d = −5. The sequence is decreasing.
2
The General Term Formula

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.

General term of an arithmetic sequence tn = a + (n − 1) × d
tₙ
The value of the nth term, what you're finding (or what you're given to solve for n)
a
The first term of the sequence (t₁)
d
The common difference, how much you add each step (can be negative)

🔵 Find tₙ

Sub in a, n, and d. Calculate.

e.g. Find t₁₀ when a=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 a or d

Sub in what you know. Solve for the unknown variable.

e.g. t₁₂=47, d=4:
a+11×4=47 → a=3

⚠️ Common mistake: The formula uses (n−1) lots of d, not n lots. Why? Because the first term (n=1) already exists, you add d zero times to get t₁. You only add d once to get t₂, twice to get t₃, and so on.
3
Sum of an Arithmetic Sequence

The sum of the first n terms is called an arithmetic series. There are two equivalent formulas, use whichever is easier for the information you have.

Formula 1, Use when you know a and d

Sₙ = n/2 × (2a + (n−1)d)

Sub in a, d, and n. This always works.

Formula 2, Use when you know first and last term

Sₙ = n/2 × (a + l)

Where l = last term (tₙ). Average the first and last, then multiply by n.

💡 Why they're the same: Formula 2 is just Formula 1 simplified, once you sub in l = a + (n−1)d and expand, you get Formula 1. On an exam, use whichever one your calculator is set up for.
4
Worked Example
🎭
Cinema Seating Problem
Row 1 has 12 seats. Each row has 3 more seats than the row in front of it.

Sequence: 12, 15, 18, 21, ...  |  a = 12, d = 3

1

How many seats are in row 10?

Use tₙ = a + (n−1)d with n = 10:
t₁₀ = 12 + (10−1) × 3 = 12 + 27 = 39 seats

2

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 ✓)

3

Total seats in the first 10 rows?

Use Sₙ = n/2 × (a + l) = 10/2 × (12 + 39)
= 5 × 51 = 255 seats

Or Formula 1: S₁₀ = 10/2 × (2×12 + 9×3) = 5 × (24+27) = 5 × 51 = 255 ✓

4

If the cinema has a total of 255 seats arranged this way, how many rows are there?

From Step 3 we already know S₁₀ = 255, so there are 10 rows. On an exam you might need to solve Sₙ = 255 using the sum formula and your calculator, but always check your answer makes sense in context.

Key checks: Does n come out as a whole number? Is it positive? Does it fit the context?
5
Practice Questions

Tap to reveal the answer. Try it yourself first!

1
The sequence 5, 9, 13, 17, ... is arithmetic. State the first term, find the common difference, and write the general term formula tₙ.
Tap to reveal ▾
a = 5, d = 9 − 5 = 4

General term: tₙ = a + (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 ✓
2
An arithmetic sequence has a = 50 and d = −6. Find t₈ and determine which term is the first negative term.
Tap to reveal ▾
t₈ = 50 + 7×(−6) = 50 − 42 = 8

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 ✓)
3
The 4th term of an arithmetic sequence is 22 and the 9th term is 47. Find the common difference and the first term.
Tap to reveal ▾
From t₄ to t₉ is 5 gaps of d: 47 − 22 = 25, so d = 25 ÷ 5 = 5

Use t₄ = 22 to find a:
a + 3d = 22 → a + 15 = 22 → a = 7

Check: 7, 12, 17, 22 ✓ (t₄=22)   t₉ = 7 + 8×5 = 47 ✓
4
Find the sum of the first 20 terms of the arithmetic sequence 6, 11, 16, 21, ...
Tap to reveal ▾
a = 6, d = 5, n = 20

Method 1 (Formula 1):
S₂₀ = 20/2 × (2×6 + 19×5) = 10 × (12 + 95) = 10 × 107 = 1070

Method 2 (Formula 2):
t₂₀ = 6 + 19×5 = 101
S₂₀ = 20/2 × (6 + 101) = 10 × 107 = 1070 ✓
5
A plumber charges $80 call-out fee plus $35 for each hour worked. Write the total charge as an arithmetic sequence (t₁ = first hour, t₂ = second hour, etc.) and find the charge for a 6-hour job.
Tap to reveal ▾
After 1 hour: $80 + $35 = $115 → a = 115
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.
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