Special Sequences & Geometric Progressions

1. Special Number Sequences

Examiners love testing if you can recognize these sequences just by looking at them. You should memorize the first few terms of each.

Square Numbers

Created by multiplying a number by itself ($n^2$).

$1, 4, 9, 16, 25, 36, 49, 64, 81, 100 \dots$

Cube Numbers

Created by multiplying a number by itself three times ($n^3$).

$1, 8, 27, 64, 125 \dots$

Triangular Numbers

These are the number of dots needed to form an equilateral triangle. You create them by adding 2, then adding 3, then adding 4, etc. The $n^{th}$ term formula is $\frac{n(n+1)}{2}$.

$1, 3, 6, 10, 15, 21 \dots$

2. Geometric Progressions (G.P.)

In a linear sequence, you add or subtract the same number every time. In a **geometric progression**, you multiply or divide by the same number every time. We call this multiplier the Common Ratio ($r$).

Example 1: Finding the next terms of a geometric sequence
Find the next two terms in the sequence: $2, 6, 18, 54 \dots$
Step 1: Find the common ratio ($r$) How do we get from 2 to 6? We multiply by 3. Check the next one: $6 \times 3 = 18$. The common ratio $r = 3$.
Step 2: Multiply to find the next terms $54 \times 3 = 162$
$162 \times 3 = 486$
Answer: 162, 486
The $n^{th}$ Term of a Geometric Sequence:
The formula is: $$ar^{n-1}$$ Note: You raise $r$ to the power of $(n-1)$, you DO NOT multiply $a \times r$ first! (Remember BIDMAS: Indices before Multiplication).
Example 2: Finding the $n^{th}$ term expression
Find the $n^{th}$ term of the sequence $5, 10, 20, 40 \dots$
Step 1: Identify $a$ and $r$ The first term is 5, so $a = 5$.
The sequence is doubling every time, so the common ratio $r = 2$.
Step 2: Plug into the formula $$ar^{n-1} = 5 \times 2^{n-1}$$ Answer: $5 \times 2^{n-1}$

3. Fibonacci-Style Sequences

In a Fibonacci sequence, you add the two previous terms together to get the next term.

The famous standard Fibonacci sequence starts with 1, 1:
$1, 1, 2, 3, 5, 8, 13, 21 \dots$

Example 3: Algebraic Fibonacci
A Fibonacci sequence starts with the terms $x$ and $y$. Write expressions for the next three terms.
Step 1: Add the first two terms 3rd term = $x + y$
Step 2: Add the 2nd and 3rd terms 4th term = $y + (x + y) = x + 2y$
Step 3: Add the 3rd and 4th terms 5th term = $(x + y) + (x + 2y) = 2x + 3y$
Answer: $x+y$, $x+2y$, $2x+3y$

10 Practice Questions & Solutions

Try these on paper, showing your steps. Click the question to check your working!

Q1: Identify the type of sequence: $1, 8, 27, 64 \dots$
These are numbers multiplied by themselves three times ($1^3, 2^3, 3^3, 4^3$).
Answer: Cube Numbers
Q2: Find the next two terms of the triangular sequence: $10, 15, 21 \dots$
Look at the gaps:
$10 \rightarrow 15$ (+5)
$15 \rightarrow 21$ (+6)
The next gap must be +7. So, $21 + 7 = 28$.
The next gap must be +8. So, $28 + 8 = 36$.
Answer: 28, 36
Q3: Find the next two terms of the geometric sequence: $80, 40, 20 \dots$
The sequence is halving every time. The common ratio $r = 0.5$ (or $\frac{1}{2}$).
$20 \times 0.5 = 10$
$10 \times 0.5 = 5$
Answer: 10, 5
Q4: A sequence has an $n^{th}$ term of $n^2 + 3$. Find the first 3 terms.
Substitute $n = 1$: $1^2 + 3 = 4$
Substitute $n = 2$: $2^2 + 3 = 7$
Substitute $n = 3$: $3^2 + 3 = 12$
Answer: 4, 7, 12
Q5: Find the $n^{th}$ term of the geometric sequence: $3, 12, 48, 192 \dots$
First term ($a$) = 3.
Common ratio ($r$): $12 \div 3 = 4$. So $r = 4$.
Formula: $ar^{n-1}$
Answer: $3 \times 4^{n-1}$
Q6: A Fibonacci-style sequence starts $3, 7 \dots$ Find the 5th term.
1st: 3
2nd: 7
3rd: $3 + 7 = 10$
4th: $7 + 10 = 17$
5th: $10 + 17 = 27$
Answer: 27
Q7: The $n^{th}$ term of a geometric sequence is $2 \times 5^{n-1}$. What is the 4th term?
Substitute $n = 4$ into the formula:
$2 \times 5^{4-1}$
$= 2 \times 5^3$
$= 2 \times 125$
Answer: 250
Q8: Find the common ratio ($r$) of the sequence: $400, 100, 25, 6.25 \dots$
To find the ratio, divide any term by the term before it.
$100 \div 400 = 0.25$ (or $\frac{1}{4}$)
Answer: $r = 0.25$
Q9: Word Problem: A population of 100 bacteria doubles every hour. How many bacteria are there after 5 hours?
This is a geometric progression.
Hour 0 (Start): 100
Hour 1: 200
Hour 2: 400
Hour 3: 800
Hour 4: 1600
Hour 5: 3200
Alternatively, using the formula: $100 \times 2^5 = 100 \times 32 = 3200$.
Answer: 3,200 bacteria
Q10: Challenge! An algebraic Fibonacci sequence starts $a, b \dots$ If the 4th term is 19 and the 5th term is 31, find the values of $a$ and $b$.
Let's build the sequence algebraically:
1st: $a$
2nd: $b$
3rd: $a + b$
4th: $a + 2b$
5th: $2a + 3b$

Now set up simultaneous equations:
(1) $a + 2b = 19$
(2) $2a + 3b = 31$

Multiply (1) by 2: $2a + 4b = 38$
Subtract (2) from this new equation:
$(2a - 2a) + (4b - 3b) = 38 - 31$
$b = 7$

Sub $b=7$ back into equation (1):
$a + 2(7) = 19 \Rightarrow a + 14 = 19 \Rightarrow a = 5$
Answer: $a = 5, b = 7$