Welcome to Sequences!
Have you ever noticed patterns in everyday life, like the rhythm of a song, the way petals grow on a sunflower, or how your savings increase by the exact same amount each week? In mathematics, when we arrange numbers in an ordered list that follows a specific rule, we call it a sequence.
Sequences are an essential part of the GCSE Mathematics syllabus. They connect number patterns directly to algebra. In this guide, we will break down sequences step-by-step so that you can tackle any exam question with confidence!
1. What is a Sequence? Key Terms to Know
Before diving into calculations, let's learn the basic vocabulary:
• Sequence: An ordered list of numbers formed by a mathematical rule. Example: \(3, 7, 11, 15, \dots\)
• Term: Each individual number in a sequence. In the sequence above, the first term is \(3\), and the third term is \(11\).
• Term-to-Term Rule: Tells you how to get from one term to the very next term (for example, "add \(4\)" or "multiply by \(2\)").
• Position-to-Term Rule (\(n^{\text{th}}\) Term): An algebraic formula that lets you calculate the value of any term using its position number \(n\).
Did you know? The three little dots \((\dots)\) at the end of a sequence are called an ellipsis. They show that the pattern continues forever into infinity!
Key Takeaway: The letter \(n\) always stands for the position number (\(n = 1\) is the \(1^{\text{st}}\) term, \(n = 2\) is the \(2^{\text{nd}}\) term, \(n = 100\) is the \(100^{\text{th}}\) term).
2. Linear (Arithmetic) Sequences
A linear sequence (sometimes called an arithmetic progression) is a sequence where the difference between consecutive terms is always the same constant number. We call this the common difference.
Finding the Next Terms using the Term-to-Term Rule
Look at the sequence: \(5, 9, 13, 17, \dots\)
To find the next term, look at the difference between the terms:
\(9 - 5 = +4\)
\(13 - 9 = +4\)
\(17 - 13 = +4\)
The term-to-term rule is "add \(4\)". The next two terms are \(17 + 4 = 21\) and \(21 + 4 = 25\).
Finding the \(n^{\text{th}}\) Term of a Linear Sequence
What if an exam question asks for the \(100^{\text{th}}\) term? Adding \(4\) one hundred times would take way too long! This is why we find the \(n^{\text{th}}\) term formula, written in the form:
\(an + b\)
Step-by-Step Method:
Step 1: Find the common difference between consecutive terms. This number is \(a\) (the multiplier in front of \(n\)).
Step 2: Multiply \(n\) by this common difference to write the first part of your rule (\(an\)).
Step 3: Compare the \(an\) times table with your actual sequence to find the adjustment value \(b\) (ask yourself: "What do I add or subtract to get from the times table to my sequence?").
Worked Example 1: Increasing Sequence
Find the \(n^{\text{th}}\) term of the sequence: \(6, 11, 16, 21, 26, \dots\)
Step 1: Find the difference: \(11 - 6 = +5\), \(16 - 11 = +5\). The common difference is \(+5\), so the rule begins with \(5n\).
Step 2: Write out the \(5\) times table (\(5n\)) under the sequence:
For \(n = 1\): \(5(1) = 5\)
For \(n = 2\): \(5(2) = 10\)
For \(n = 3\): \(5(3) = 15\)
Step 3: How do you get from the \(5n\) row to the actual sequence?
From \(5\) to \(6\): \(+1\)
From \(10\) to \(11\): \(+1\)
From \(15\) to \(16\): \(+1\)
Therefore, the \(n^{\text{th}}\) term is \(5n + 1\).
Worked Example 2: Decreasing Sequence
Find the \(n^{\text{th}}\) term of the sequence: \(20, 17, 14, 11, \dots\)
Step 1: Find the difference: each term goes down by \(3\), so the difference is \(-3\). This gives us \(-3n\).
Step 2: Test \(n = 1\): \(-3 \times 1 = -3\).
Step 3: How do you get from \(-3\) to the first term (\(20\))? You add \(23\) because \(-3 + 23 = 20\).
Therefore, the \(n^{\text{th}}\) term is \(-3n + 23\) (or written as \(23 - 3n\)).
Memory Trick (The "Zero Term" Shortcut):
The \(n^{\text{th}}\) term is always \(\text{Difference} \times n + \text{Term } 0\).
To find Term 0, simply step backwards from the first term by reversing the rule!
Using the \(n^{\text{th}}\) Term to Solve Problems
1. Finding a specific term:
Question: Find the \(50^{\text{th}}\) term of the sequence with \(n^{\text{th}}\) term \(4n - 3\).
Solution: Substitute \(n = 50\) into the rule: \(4(50) - 3 = 200 - 3 = 197\).
2. Checking if a number is in a sequence:
Question: Is \(87\) a term in the sequence \(4n + 3\)?
Solution: Set the formula equal to \(87\) and solve for \(n\):
\(4n + 3 = 87\)
\(4n = 84\)
\(n = \frac{84}{4} = 21\)
Since \(n = 21\) is a whole number (an integer), yes, \(87\) is the \(21^{\text{st}}\) term of the sequence!
Key Takeaway: For linear sequences, the first difference is constant. The formula is always \(dn + c\), where \(d\) is the common difference.
3. Special Sequences to Recognise
Not every sequence goes up by the same amount each time. Here are famous sequences you must be able to identify:
1. Square Numbers (\(n^2\)):
\(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \dots\)
Formed by multiplying a number by itself: \(1^2, 2^2, 3^2, 4^2, \dots\)
2. Cube Numbers (\(n^3\)):
\(1, 8, 27, 64, 125, \dots\)
Formed by multiplying a number by itself three times: \(1^3, 2^3, 3^3, \dots\)
3. Triangular Numbers:
\(1, 3, 6, 10, 15, 21, \dots\)
These numbers represent dots that can form an equilateral triangle. The rule is: add \(2\), then add \(3\), then add \(4\), then add \(5\), and so on.
4. The Fibonacci Sequence:
\(1, 1, 2, 3, 5, 8, 13, 21, 34, \dots\)
Rule: Each term is found by adding together the two previous terms.
\(1 + 1 = 2\)
\(1 + 2 = 3\)
\(2 + 3 = 5\)
\(3 + 5 = 8\)
5. Geometric Sequences (Multiplying / Dividing):
Example: \(3, 6, 12, 24, 48, \dots\)
Instead of adding a number, you multiply by a constant factor each time (here, "multiply by \(2\)").
Key Takeaway: When you see a pattern where the differences are not constant, always check if it's related to square numbers, powers, or adding previous terms (Fibonacci).
4. Visual and Pattern Sequences
Exam questions often show patterns made of matchsticks, tiles, or dots.
How to tackle pattern problems:
Step 1: Count the items in Pattern \(1\), Pattern \(2\), and Pattern \(3\).
Step 2: Write the counts down as a regular number sequence.
Step 3: Find the \(n^{\text{th}}\) term formula using your standard linear sequence method.
Step 4: Use the formula to answer the question.
Example:
Pattern 1 has \(4\) matchsticks.
Pattern 2 has \(7\) matchsticks.
Pattern 3 has \(10\) matchsticks.
The sequence of numbers is \(4, 7, 10, \dots\)
Difference \(= +3 \implies 3n\).
When \(n = 1\), \(3(1) + 1 = 4\).
Formula \(= 3n + 1\).
How many matches are in Pattern \(20\)? Just substitute \(n = 20\): \(3(20) + 1 = 61\) matchsticks.
5. Quadratic Sequences
A quadratic sequence is one where the first differences change, but the second differences (the difference between the differences) are constant. The general formula for a quadratic sequence is:
\(an^2 + bn + c\)
Simple Quadratic Sequences (\(n^2 + c\))
Look at: \(3, 6, 11, 18, 27, \dots\)
Let's compare this with the standard square numbers \(n^2\):
Square numbers (\(n^2\)): \(1, 4, 9, 16, 25\)
Our sequence: \(3, 6, 11, 18, 27\)
Notice that every single term in our sequence is exactly \(2\) greater than \(n^2\).
Therefore, the \(n^{\text{th}}\) term is simply \(n^2 + 2\).
General Method for Finding the \(n^{\text{th}}\) Term of \(an^2 + bn + c\)
Golden Rule: The coefficient of \(n^2\) (the value of \(a\)) is always half of the second difference.
Worked Example:
Find the \(n^{\text{th}}\) term of: \(4, 11, 22, 37, 56, \dots\)
Step 1: Find the first and second differences:
Terms: \(4, \quad 11, \quad 22, \quad 37, \quad 56\)
\(1^{\text{st}}\) differences: \(+7, \quad +11, \quad +15, \quad +19\)
\(2^{\text{nd}}\) differences: \(+4, \quad +4, \quad +4\)
Step 2: Find \(a\):
\(a = \frac{\text{Second Difference}}{2} = \frac{4}{2} = 2\).
This means our formula contains \(2n^2\).
Step 3: Subtract \(2n^2\) from the original sequence:
Original sequence: \(4, 11, 22, 37, 56\)
Calculate \(2n^2\):
For \(n=1\): \(2(1^2) = 2\)
For \(n=2\): \(2(2^2) = 8\)
For \(n=3\): \(2(3^2) = 18\)
For \(n=4\): \(2(4^2) = 32\)
Subtract \(( \text{Sequence} - 2n^2 )\):
\(4 - 2 = 2\)
\(11 - 8 = 3\)
\(22 - 18 = 4\)
\(37 - 32 = 5\)
Step 4: Find the linear rule for the remaining sequence (\(2, 3, 4, 5, \dots\)):
This is a simple linear sequence going up by \(1\), so its rule is \(1n + 1\) (or \(n + 1\)).
Step 5: Combine everything together:
\(n^{\text{th}} \text{ term} = 2n^2 + n + 1\)
Key Takeaway: If the second difference is constant, it is a quadratic sequence. Halve the second difference to get \(a\) in \(an^2\).
6. Common Mistakes to Avoid
• Confusing \(n\) with the term value: Remember that \(n\) is the position number (\(1, 2, 3, \dots\)), not the number in the sequence itself.
• Negative signs in decreasing sequences: If the sequence goes down, the coefficient of \(n\) must be negative (e.g., \(-4n\)). Don't accidentally write \(+4n\).
• Forgetting to halve the second difference: In quadratic sequences, students often forget to divide the second difference by \(2\). Always remember: \(a = \frac{\text{second difference}}{2}\).
• Not checking your formula: Always test your final formula using \(n = 1\) and \(n = 2\) to make sure it produces the first two terms correctly!
Quick Review Summary
1. Linear Sequence: \(1^{\text{st}}\) difference is constant \(\implies an + b\).
2. Quadratic Sequence: \(2^{\text{nd}}\) difference is constant \(\implies an^2 + bn + c\).
3. Fibonacci Sequence: Add the two previous numbers together.
4. Checking membership: Set \(n^{\text{th}} \text{ term} = \text{number}\) and solve for \(n\). If \(n\) is a positive whole number, it is in the sequence!