Sequences are ordered lists of numbers that follow specific patterns. Each number in a sequence is called a term. Understanding sequences involves recognising patterns, finding the next terms, and developing formulas to determine any term in the sequence without having to list all previous terms.
At GCSE level, you need to work with various types of sequences including linear (arithmetic), quadratic, and special sequences such as Fibonacci-type sequences. Being able to recognise and work with sequences is an important skill that develops pattern recognition and algebraic thinking.
TLDR
- Linear (Arithmetic) Sequences: Add the same value each time (common difference)
- Find the nth term using formula \(an + b\) where \(a\) is the common difference
- Example: \(2, 5, 8, 11, 14…\) has nth term \(3n – 1\)
- Quadratic Sequences: Second differences are constant
- Find the nth term using the form \(an^2 + bn + c\)
- Example: \(2, 6, 12, 20, 30…\) has nth term \(n^2 + n\)
- Special Sequences:
- Triangular numbers: \(1, 3, 6, 10, 15, 21…\)
- Square numbers: \(1, 4, 9, 16, 25…\)
- Fibonacci-type: Each term is the sum of the two previous terms
- Finding Terms:
- For linear: Substitute \(n\) into the formula \(an + b\)
- For quadratic: Substitute \(n\) into the formula \(an^2 + bn + c\)
- For special sequences: Apply the specific rule for that sequence
Key Facts
🔑 Key Fact: The position of a term in a sequence is denoted by \(n\). The 1st term has \(n=1\), the 2nd term has \(n=2\), and so on.
🔑 Key Fact: For linear sequences, the nth term formula is always in the form \(an + b\), where \(a\) is the common difference between terms.
🔑 Key Fact: For quadratic sequences, the second differences are always constant, and the nth term formula is in the form \(an^2 + bn + c\).
🔑 Key Fact: When finding the nth term of a quadratic sequence, the coefficient of \(n^2\) is half of the constant second difference.
Linear Sequences (Arithmetic Sequences)
Linear sequences have a constant difference between consecutive terms. This constant difference is called the common difference.
Finding the Common Difference
To find the common difference, subtract any term from the term that follows it:
- Common difference = Next term – Current term
Finding the nth Term Formula
For a linear sequence with common difference \(a\), the nth term formula is \(an + b\) where:
- \(a\) is the common difference
- \(b\) is a constant that needs to be determined
To find \(b\), substitute \(n=1\) into the formula and set it equal to the first term in the sequence.
Example 1
Find the nth term of the sequence: \(5, 8, 11, 14, 17, …\)
Step 1: Find the common difference.
\(8 – 5 = 3\)
\(11 – 8 = 3\)
\(14 – 11 = 3\)
\(17 – 14 = 3\)
The common difference is 3, so \(a = 3\).
Step 2: Determine the formula \(an + b = 3n + b\).
Step 3: Find \(b\) by substituting \(n = 1\) for the first term.
\(3(1) + b = 5\)
\(3 + b = 5\)
\(b = 2\)
Step 4: Write the nth term formula.
\(nth\) term \(= 3n + 2\)
Step 5: Verify by checking a few terms.
\(n = 1: 3(1) + 2 = 5\) ✓
\(n = 2: 3(2) + 2 = 8\) ✓
\(n = 3: 3(3) + 2 = 11\) ✓
Therefore, the nth term is \(3n + 2\).
Example 2
Find the nth term of the sequence: \(3, 5, 7, 9, 11, …\)
Step 1: Find the common difference.
\(5 – 3 = 2\)
\(7 – 5 = 2\)
\(9 – 7 = 2\)
\(11 – 9 = 2\)
The common difference is 2, so \(a = 2\).
Step 2: Determine the formula \(an + b = 2n + b\).
Step 3: Find \(b\) by substituting \(n = 1\) for the first term.
\(2(1) + b = 3\)
\(2 + b = 3\)
\(b = 1\)
Step 4: Write the nth term formula.
\(nth\) term \(= 2n + 1\)
Step 5: Verify by checking a few terms.
\(n = 1: 2(1) + 1 = 3\) ✓
\(n = 2: 2(2) + 1 = 5\) ✓
\(n = 3: 2(3) + 1 = 7\) ✓
Therefore, the nth term is \(2n + 1\).
Example 3
Find the nth term of the sequence: \(7, 4, 1, -2, -5, …\)
Step 1: Find the common difference.
\(4 – 7 = -3\)
\(1 – 4 = -3\)
\(-2 – 1 = -3\)
\(-5 – (-2) = -3\)
The common difference is -3, so \(a = -3\).
Step 2: Determine the formula \(an + b = -3n + b\).
Step 3: Find \(b\) by substituting \(n = 1\) for the first term.
\(-3(1) + b = 7\)
\(-3 + b = 7\)
\(b = 10\)
Step 4: Write the nth term formula.
\(nth\) term \(= -3n + 10\)
Step 5: Verify by checking a few terms.
\(n = 1: -3(1) + 10 = 7\) ✓
\(n = 2: -3(2) + 10 = 4\) ✓
\(n = 3: -3(3) + 10 = 1\) ✓
Therefore, the nth term is \(-3n + 10\).
Quadratic Sequences
Quadratic sequences have a second difference that is constant. The nth term formula is in the form \(an^2 + bn + c\).
Finding the Second Difference
To find the second difference:
- Calculate the first differences (the differences between consecutive terms)
- Calculate the differences between consecutive first differences
Finding the nth Term Formula
For a quadratic sequence with a constant second difference of \(d\), the coefficient of \(n^2\) is \(\frac{d}{2}\), so \(a = \frac{d}{2}\).
To find \(b\) and \(c\):
- Create a table with the original sequence and the values of \(an^2\) for each position
- Subtract the \(an^2\) values from the original sequence to get a linear sequence
- Find the nth term of this linear sequence (\(bn + c\))
- The nth term of the original quadratic sequence is \(an^2 + bn + c\)
Example 4
Find the nth term of the sequence: \(2, 6, 12, 20, 30, …\)
Step 1: Find the first and second differences.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 2 | 6 | 12 | 20 | 30 |
| First differences | 4 | 6 | 8 | 10 | |
| Second differences | 2 | 2 | 2 |
The second difference is constant at 2, so this is a quadratic sequence.
Step 2: Find the coefficient of \(n^2\) by dividing the second difference by 2.
\(a = \frac{2}{2} = 1\)
Step 3: Create a table to find the linear part of the sequence.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 2 | 6 | 12 | 20 | 30 |
| \(n^2\) values | 1 | 4 | 9 | 16 | 25 |
| Sequence – \(n^2\) | 1 | 2 | 3 | 4 | 5 |
Step 4: The resulting sequence \(1, 2, 3, 4, 5, …\) is a linear sequence with common difference 1. Its nth term is \(n\).
Step 5: Combine the quadratic and linear parts to get the nth term.
\(nth\) term \(= n^2 + n\)
Step 6: Verify by checking a few terms.
\(n = 1: 1^2 + 1 = 2\) ✓
\(n = 2: 2^2 + 2 = 6\) ✓
\(n = 3: 3^2 + 3 = 12\) ✓
Therefore, the nth term is \(n^2 + n\).
Example 5
Find the nth term of the sequence: \(5, 8, 13, 20, 29, …\)
Step 1: Find the first and second differences.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 5 | 8 | 13 | 20 | 29 |
| First differences | 3 | 5 | 7 | 9 | |
| Second differences | 2 | 2 | 2 |
The second difference is constant at 2, so this is a quadratic sequence.
Step 2: Find the coefficient of \(n^2\) by dividing the second difference by 2.
\(a = \frac{2}{2} = 1\)
Step 3: Create a table to find the linear part of the sequence.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 5 | 8 | 13 | 20 | 29 |
| \(n^2\) values | 1 | 4 | 9 | 16 | 25 |
| Sequence – \(n^2\) | 4 | 4 | 4 | 4 | 4 |
Step 4: The resulting sequence \(4, 4, 4, 4, 4, …\) is a constant sequence. Its nth term is 4.
Step 5: Combine the quadratic and constant parts to get the nth term.
\(nth\) term \(= n^2 + 4\)
Step 6: Verify by checking a few terms.
\(n = 1: 1^2 + 4 = 5\) ✓
\(n = 2: 2^2 + 4 = 8\) ✓
\(n = 3: 3^2 + 4 = 13\) ✓
Therefore, the nth term is \(n^2 + 4\).
Special Sequences
Triangular Numbers
Triangular numbers represent the number of dots needed to form an equilateral triangle. The sequence is \(1, 3, 6, 10, 15, 21, …\)
The formula for the nth triangular number is:
\(T_n = \frac{n(n+1)}{2}\)
Example 6
Find the 8th triangular number.
Step 1: Use the formula \(T_n = \frac{n(n+1)}{2}\) with \(n = 8\).
Step 2: Calculate the result.
\(T_8 = \frac{8(8+1)}{2} = \frac{8 \times 9}{2} = \frac{72}{2} = 36\)
Therefore, the 8th triangular number is 36.
Square Numbers
Square numbers are the result of squaring integers. The sequence is \(1, 4, 9, 16, 25, 36, …\)
The formula for the nth square number is:
\(S_n = n^2\)
Cubic Numbers
Cubic numbers are the result of cubing integers. The sequence is \(1, 8, 27, 64, 125, 216, …\)
The formula for the nth cubic number is:
\(C_n = n^3\)
Fibonacci-type Sequences
In Fibonacci-type sequences, each term is the sum of the two previous terms. The classic Fibonacci sequence starts with 1, 1 and continues as \(1, 1, 2, 3, 5, 8, 13, 21, …\)
To find the next term, add the two previous terms:
\(F_n = F_{n-1} + F_{n-2}\)
Geometric Sequences (Higher Tier)
In geometric sequences, each term is found by multiplying the previous term by a constant value called the common ratio. The sequence has the form \(a, ar, ar^2, ar^3, …\)
The formula for the nth term is:
\(G_n = ar^{n-1}\)
where \(a\) is the first term and \(r\) is the common ratio.
Example 7
Find the 6th term of the geometric sequence \(3, 6, 12, 24, …\)
Step 1: Find the common ratio by dividing consecutive terms.
\(\frac{6}{3} = 2\)
\(\frac{12}{6} = 2\)
\(\frac{24}{12} = 2\)
The common ratio is 2.
Step 2: Use the formula \(G_n = ar^{n-1}\) with \(a = 3\), \(r = 2\), and \(n = 6\).
Step 3: Calculate the result.
\(G_6 = 3 \times 2^{6-1} = 3 \times 2^5 = 3 \times 32 = 96\)
Therefore, the 6th term is 96.
Common Mistakes to Avoid
- Confusing the term number with the term value:
– Incorrect: Thinking that for the sequence \(3, 6, 9, 12, …\), the first term is \(n=3\).
– Correct: \(n=1\) corresponds to the first term, which is 3. The sequence is \(3n\). - Incorrect identification of sequence type:
– Incorrect: Treating a quadratic sequence as linear.
– Correct: Calculate the second differences to check if the sequence is quadratic. - Forgetting to verify the formula:
– Incorrect: Finding the formula but not checking if it works for the given terms.
– Correct: Always verify your formula by calculating several terms and comparing with the original sequence. - Errors in finding the constant term:
– Incorrect: For the sequence \(2, 5, 8, 11, …\), writing the formula as \(3n\).
– Correct: The formula is \(3n – 1\) because the first term requires \(3(1) – 1 = 2\). - Incorrectly calculating second differences:
– Incorrect: Calculating first differences correctly but then adding instead of subtracting to find second differences.
– Correct: Second differences are found by subtracting consecutive first differences.
When you see a practice question, you’ll learn much more if you attempt it before you look at the solution. Use the solution to check your answer, or if you get stuck.
Questions
Try these questions to practice your understanding of sequences:
- Find the next two terms in the sequence: \(5, 9, 13, 17, …\)
- Find the nth term formula for the sequence: \(4, 9, 14, 19, 24, …\)
- Find the 15th term of the sequence with nth term formula: \(2n – 3\)
- Find the nth term formula for the quadratic sequence: \(3, 8, 15, 24, 35, …\)
- Find the 10th term in the Fibonacci-type sequence: \(3, 5, 8, 13, 21, …\)
- Find the missing terms in the sequence: \(4, 9, \_, \_, 49, \_, 121\)
- Higher tier: Find the nth term formula for the quadratic sequence: \(1, 6, 15, 28, 45, …\)
- Higher tier: Find the 8th term of the geometric sequence with first term 5 and common ratio 3.
Solutions
Question 1
Find the next two terms in the sequence: \(5, 9, 13, 17, …\)
Step 1: Find the common difference.
\(9 – 5 = 4\)
\(13 – 9 = 4\)
\(17 – 13 = 4\)
The common difference is 4.
Step 2: Calculate the next two terms by adding 4 to each preceding term.
Next term: \(17 + 4 = 21\)
Next term after that: \(21 + 4 = 25\)
Therefore, the next two terms are 21 and 25.
Question 2
Find the nth term formula for the sequence: \(4, 9, 14, 19, 24, …\)
Step 1: Find the common difference.
\(9 – 4 = 5\)
\(14 – 9 = 5\)
\(19 – 14 = 5\)
\(24 – 19 = 5\)
The common difference is 5, so \(a = 5\).
Step 2: Determine the formula \(an + b = 5n + b\).
Step 3: Find \(b\) by substituting \(n = 1\) for the first term.
\(5(1) + b = 4\)
\(5 + b = 4\)
\(b = -1\)
Step 4: Write the nth term formula.
\(nth\) term \(= 5n – 1\)
Step 5: Verify by checking a few terms.
\(n = 1: 5(1) – 1 = 4\) ✓
\(n = 2: 5(2) – 1 = 9\) ✓
\(n = 3: 5(3) – 1 = 14\) ✓
Therefore, the nth term is \(5n – 1\).
Question 3
Find the 15th term of the sequence with nth term formula: \(2n – 3\)
Step 1: Substitute \(n = 15\) into the formula.
\(2(15) – 3 = 30 – 3 = 27\)
Therefore, the 15th term is 27.
Question 4
Find the nth term formula for the quadratic sequence: \(3, 8, 15, 24, 35, …\)
Step 1: Find the first and second differences.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 3 | 8 | 15 | 24 | 35 |
| First differences | 5 | 7 | 9 | 11 | |
| Second differences | 2 | 2 | 2 |
The second difference is constant at 2, so this is a quadratic sequence.
Step 2: Find the coefficient of \(n^2\) by dividing the second difference by 2.
\(a = \frac{2}{2} = 1\)
Step 3: Create a table to find the linear part of the sequence.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 3 | 8 | 15 | 24 | 35 |
| \(n^2\) values | 1 | 4 | 9 | 16 | 25 |
| Sequence – \(n^2\) | 2 | 4 | 6 | 8 | 10 |
Step 4: The resulting sequence \(2, 4, 6, 8, 10, …\) is a linear sequence with common difference 2. Its nth term is \(2n\).
Step 5: Combine the quadratic and linear parts to get the nth term.
\(nth\) term \(= n^2 + 2n\)
Step 6: Verify by checking a few terms.
\(n = 1: 1^2 + 2(1) = 1 + 2 = 3\) ✓
\(n = 2: 2^2 + 2(2) = 4 + 4 = 8\) ✓
\(n = 3: 3^2 + 2(3) = 9 + 6 = 15\) ✓
Therefore, the nth term is \(n^2 + 2n\).
Question 5
Find the 10th term in the Fibonacci-type sequence: \(3, 5, 8, 13, 21, …\)
Step 1: Continue the sequence by adding consecutive terms.
\(F_6 = F_5 + F_4 = 21 + 13 = 34\)
\(F_7 = F_6 + F_5 = 34 + 21 = 55\)
\(F_8 = F_7 + F_6 = 55 + 34 = 89\)
\(F_9 = F_8 + F_7 = 89 + 55 = 144\)
\(F_{10} = F_9 + F_8 = 144 + 89 = 233\)
Therefore, the 10th term is 233.
Question 6
Find the missing terms in the sequence: \(4, 9, \_, \_, 49, \_, 121\)
Step 1: The terms appear to be square numbers, so let’s check.
\(4 = 2^2\)
\(9 = 3^2\)
\(49 = 7^2\)
\(121 = 11^2\)
Step 2: The sequence follows the pattern of squares of: \(2, 3, \_, \_, 7, \_, 11\)
Step 3: To complete the sequence, we need the terms between 3 and 7, and between 7 and 11.
Between 3 and 7: 4, 5, 6
Between 7 and 11: 8, 9, 10
Step 4: The missing squares are:
\(4^2 = 16\)
\(5^2 = 25\)
\(9^2 = 81\)
Therefore, the complete sequence is \(4, 9, 16, 25, 49, 81, 121\).
Question 7 (Higher tier)
Find the nth term formula for the quadratic sequence: \(1, 6, 15, 28, 45, …\)
Step 1: Find the first and second differences.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 1 | 6 | 15 | 28 | 45 |
| First differences | 5 | 9 | 13 | 17 | |
| Second differences | 4 | 4 | 4 |
The second difference is constant at 4, so this is a quadratic sequence.
Step 2: Find the coefficient of \(n^2\) by dividing the second difference by 2.
\(a = \frac{4}{2} = 2\)
Step 3: Create a table to find the linear part of the sequence.
| Term number | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | 1 | 6 | 15 | 28 | 45 |
| \(2n^2\) values | 2 | 8 | 18 | 32 | 50 |
| Sequence – \(2n^2\) | -1 | -2 | -3 | -4 | -5 |
Step 4: The resulting sequence \(-1, -2, -3, -4, -5, …\) is a linear sequence with common difference -1. Its nth term is \(-n\).
Step 5: Combine the quadratic and linear parts to get the nth term.
\(nth\) term \(= 2n^2 – n\)
Step 6: Verify by checking a few terms.
\(n = 1: 2(1)^2 – 1 = 2 – 1 = 1\) ✓
\(n = 2: 2(2)^2 – 2 = 8 – 2 = 6\) ✓
\(n = 3: 2(3)^2 – 3 = 18 – 3 = 15\) ✓
Therefore, the nth term is \(2n^2 – n\).
Question 8 (Higher tier)
Find the 8th term of the geometric sequence with first term 5 and common ratio 3.
Step 1: Use the formula \(G_n = ar^{n-1}\) with \(a = 5\), \(r = 3\), and \(n = 8\).
Step 2: Calculate the result.
\(G_8 = 5 \times 3^{8-1} = 5 \times 3^7 = 5 \times 2187 = 10935\)
Therefore, the 8th term is 10935.
Summary
- Linear (Arithmetic) Sequences have a constant common difference between consecutive terms. The nth term formula is \(an + b\), where \(a\) is the common difference.
- Quadratic Sequences have a constant second difference. The nth term formula is \(an^2 + bn + c\), where \(a\) is half the constant second difference.
- Special Sequences include triangular numbers, square numbers, cubic numbers, and Fibonacci-type sequences, each with their own pattern and formula.
- To find the nth term:
- Identify the type of sequence by checking differences
- For linear: determine the common difference and constant term
- For quadratic: find the coefficient of \(n^2\) and analyse the remaining linear part
- For special sequences: apply the specific formula for that type
Remember, recognising patterns is the key to working with sequences. Practice identifying sequence types and finding nth term formulas to build confidence in this important area of mathematics.
By Atomic