Triangular, Square and Cube Numbers and Arithmetic Progressions Revision Notes | GCSE AQA Higher Maths

1. Why Number Sequences Matter

Sequences are everywhere — from the spiral of a nautilus shell to the way interest compounds in a bank account. In GCSE Maths, sequences questions appear on both questions without a calculator and calculator-allowed questions, and they reward students who can spot patterns quickly and write them down precisely.

This lesson covers three special families of numbers — triangular, square, and cube — plus the workhorse of all sequences: the arithmetic progression. Master these and you have a toolkit that handles the vast majority of sequence questions the examiner can throw at you.

2. Square and Cube Numbers

Start with something you almost certainly already know. When you multiply a whole number by itself, you get a square number. The name comes from geometry: $3 \times 3 = 9$ dots can be arranged into a perfect $3 \times 3$ square.

Cube numbers extend the same idea into three dimensions. Instead of a flat square of dots, picture a cube of unit blocks: $3 \times 3 \times 3 = 27$ small cubes fill a larger cube with side length 3.

Memorising this table up to $n = 10$ pays dividends across many topics — from surds to volume calculations. Notice that 64 appears in both columns: $4^3 = 64$ and $8^2 = 64$. The examiner occasionally exploits this overlap.

3. Triangular Numbers

Triangular numbers are slightly less familiar, but the pattern is beautiful once you see it. Imagine stacking rows of dots: one dot on top, two below, three below that, and so on. Each total gives the next triangular number.

The sequence of triangular numbers is: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,... The differences between consecutive terms are 2, 3, 4, 5, 6,... — each one larger than the last. This increasing difference is the hallmark of a triangular sequence.

Here is a neat way to see why this formula works. Write the numbers 1 to $n$ forwards, then write them backwards underneath. Every pair adds to $(n+1)$, and there are $n$ pairs, giving a total of $n(n+1)$. Since you used each number twice, divide by 2. That is the formula — no memorisation needed if you understand the reasoning.

Worked example: Finding the 8th Triangular Number

Find the 8th triangular number using the formula $T_n = \frac{n(n+1)}{2}$.

1. $n = 8$ — We want the 8th term, so n = 8. [M1 — correct substitution]
2. $T_8 = \frac{8 \times (8 + 1)}{2}$ — Replace n with 8 in $\frac{n(n+1)}{2}$.
3. $T_8 = \frac{8 \times 9}{2}$ — $8 + 1 = 9$. Keep the fraction — do not round mid-calculation.
4. $T_8 = \frac{72}{2}$ — $8 \times 9 = 72$. Verify: $8 \times 9 = 8 \times 10 - 8 = 80 - 8 = 72$. Correct.
5. $T_8 = 36$ — $72 \div 2 = 36$. [A1 — correct answer] Notice that 36 also appears in the square number sequence ($6^2 = 36$) — another overlap the examiner might exploit.

Answer: $T_8 = 36$

4. Arithmetic Progressions

Square and triangular sequences have differences that change. An arithmetic progression (AP) is different: its differences are constant. Every term is obtained by adding the same fixed amount to the previous term. That fixed amount is called the common difference.

Examples of arithmetic progressions: $3, 7, 11, 15, 19, \ldots$ (first term $a = 3$, common difference $d = 4$) and $20, 17, 14, 11, 8, \ldots$ (first term $a = 20$, common difference $d = -3$). Notice that $d$ can be negative — the sequence then decreases.

The formula $a_n = a + (n-1)d$ works because to reach the nth term from the first, you add $d$ exactly $(n-1)$ times — not $n$ times. This is the most common error students make, so it is worth pausing on.

Worked example: Finding the 20th Term of an Arithmetic Progression

An arithmetic progression has first term 7 and common difference 5. Find the 20th term.

1. $a = 7,\quad d = 5,\quad n = 20$ — Read off the given values before substituting. [M1 — correct identification]
2. $a_n = a + (n-1)d$ — State the formula explicitly — this earns a method mark even if arithmetic slips later.
3. $a_{20} = 7 + (20 - 1) \times 5$ — Replace $a$, $n$, and $d$ with the given numbers. [M1 — correct substitution]
4. $a_{20} = 7 + 19 \times 5$ — $20 - 1 = 19$. Do the bracket before multiplying.
5. $a_{20} = 7 + 95$ — $19 \times 5 = 95$. Verify: $20 \times 5 = 100$, then $100 - 5 = 95$. Correct.
6. $a_{20} = 102$ — $7 + 95 = 102$. [A1 — correct final answer]

Answer: $a_{20} = 102$

The formula also works in reverse. If you know a term and its position, you can find $d$ or $a$. This is where the examiner adds a twist — they give you two terms and ask you to find the first term or the common difference.

Worked example: Finding the First Term and Common Difference

In an arithmetic progression, the 4th term is 19 and the 7th term is 31. Find the first term and the common difference.

1. $a + 3d = 19 \quad \text{(term 4)}$ $a + 6d = 31 \quad \text{(term 7)}$ — Use $a_n = a + (n-1)d$. For $n=4$: $a + 3d = 19$. For $n=7$: $a + 6d = 31$. [M1 — setting up both equations]
2. $(a + 6d) - (a + 3d) = 31 - 19$ $3d = 12$ — Subtracting eliminates $a$, leaving only $d$. This is simultaneous equations in disguise.
3. $d = \frac{12}{3} = 4$ — $12 \div 3 = 4$. The common difference is 4. [A1]
4. $a + 3 \times 4 = 19$ $a + 12 = 19$ $a = 7$ — Use the first equation. $3 \times 4 = 12$, then $19 - 12 = 7$. [A1]
5. $a + 6d = 7 + 6 \times 4 = 7 + 24 = 31$ ✓ — $6 \times 4 = 24$, $7 + 24 = 31$. This matches the given 7th term, confirming both answers are correct.

Answer: First term $a = 7$, common difference $d = 4$

5. Recognising Sequence Types

The examiner will often give you a sequence and ask you to identify its type or find a missing term. Here is a reliable decision process: look at the differences between consecutive terms first.

• Differences are constant (e.g. all +4): arithmetic progression
• Differences increase by 1 each time (e.g. +2, +3, +4, +5): triangular numbers
• Differences are 3, 5, 7, 9,... (odd numbers increasing by 2): square numbers
• Terms are 1, 8, 27, 64, 125: cube numbers — recognise these directly from the table
• Differences are not constant and do not fit the above patterns: may be geometric or quadratic (beyond this lesson)

For square numbers, the differences between consecutive terms are always consecutive odd numbers: $4-1=3$, $9-4=5$, $16-9=7$, $25-16=9$, and so on. This is a reliable fingerprint for the square number sequence.

6. Practice Problems

Now that all four sequence types are in your toolkit, it is time to practise applying them under exam conditions. Work through each problem before checking the solution — the attempt is where the learning happens.

7. Exam Technique and Common Mistakes

Sequence questions are reliable mark-earners — but only if you avoid the traps that cost students marks every year. The mistakes below are drawn from real examiner reports, so they are worth taking seriously.

One final examiner insight: questions sometimes combine sequence types. For example, you might be asked whether a particular square number is also a triangular number, or to find the arithmetic progression formed by every other triangular number. Approach these by applying each formula separately and then comparing results.

The graph above plots the square number formula $y = n^2$ and the triangular number formula $y = \frac{n(n+1)}{2}$ on the same axes. Both curves grow, but the square numbers grow faster. At $n = 8$, the square number is 64 while the triangular number is 36 — a gap that widens as $n$ increases. Visualising this helps you sense-check answers: for n greater than 1, a triangular number should always be smaller than the corresponding square number for the same $n$.