CIE IGCSE Mathematics (0580) Sequences Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Sequences Study Notes
LEARNING OBJECTIVE
- Understanding the term-to-term rule and relationships between different sequences.
Key Concepts:
- Sequences
- nth Term of a Sequence
Sequences: Continuing Patterns
Sequences: Continuing Patterns
A sequence is a list of numbers arranged in a specific order according to a rule or pattern. Subscript notation may be used to represent terms in a sequence. For example, \( T_1 \) is the first term, \( T_2 \) is the second term, and \( T_n \) is the \( n^{\text{th}} \) term of the sequence \( T \).
Types of Sequences:
- Arithmetic Sequence: A sequence where each term is obtained by adding or subtracting the same value (called the common difference).
- Geometric Sequence: A sequence where each term is obtained by multiplying or dividing by the same number (called the common ratio).
- Other Patterns: Includes square numbers, cube numbers, Fibonacci sequences, etc.
Nth Term Formulas:
- Arithmetic Sequence: \( T_n = a + (n – 1)d \), where \( a \) is the first term and \( d \) is the common difference.
- Geometric Sequence: \( T_n = ar^{n – 1} \), where \( a \) is the first term and \( r \) is the common ratio.
How to Continue a Sequence:
- Look at the difference or ratio between terms.
- Identify the rule being followed.
- Apply the rule to extend the sequence.
Example:
Continue the sequence: 4, 7, 10, 13, …
▶️ Answer/Explanation
Common difference = 3
\( T_n = 4 + (n – 1)\times 3 = 3n + 1 \)
Next terms: 16, 19, 22
Example:
Continue the sequence: 2, 6, 18, 54, …
▶️ Answer/Explanation
Common ratio = 3
\( T_n = 2 \times 3^{n – 1} \)
Next terms: 162, 486
Example:
Continue the sequence: 1, 4, 9, 16, …
▶️ Answer/Explanation
This is a sequence of square numbers: \( 1^2, 2^2, 3^2, 4^2 \)
\( T_n = n^2 \)
Next terms: \( 25, 36 \)
Recognising Patterns in Sequences
Recognising Patterns in Sequences
Recognising patterns in sequences helps us understand how the sequence progresses. Patterns can involve adding, subtracting, multiplying, dividing, or applying more complex operations.
Types of Rules:
- Term-to-term rule: Describes how to go from one term to the next.
- Position-to-term rule: Also known as the “nth term rule” (covered later). Gives a formula to find any term in the sequence.
Common Relationships Between Sequences:
- Sequences can be related by a constant difference or ratio.
- Sometimes, two sequences have the same rule but start with different values (e.g., 2, 4, 6, 8… and 3, 5, 7, 9… both increase by 2).
- Patterns can include alternating signs, doubling, halving, or using square/cube numbers.
Example :
Identify the term-to-term rule for the sequence: 5, 8, 11, 14, …
▶️ Answer/Explanation
Add 3 to each term.
Term-to-term rule: Add 3
Example :
Compare the two sequences: A = 2, 4, 6, 8… and B = 3, 6, 9, 12…
▶️ Answer/Explanation
Sequence A increases by 2; Sequence B increases by 3.
So both are arithmetic sequences but with different common differences.
Example :
Describe the pattern: 1, -2, 4, -8, 16, …
▶️ Answer/Explanation
The absolute value doubles each time, and the sign alternates.
Term-to-term rule: Multiply by -2
Finding and Using the nth Term of a Sequence
Finding and Using the nth Term of a Sequence
What Is the nth Term?
The nth term is a general formula used to find any term in a sequence based on its position \( n \). It helps in predicting future values and understanding patterns in sequences.
(a) Linear Sequences
A linear sequence has a constant first difference.
Its nth term has the form: \( \text{nth term} = an + b \)
Where \( a \) is the common difference and \( b \) adjusts for the first term.
Example:
Find the nth term of the sequence: 3, 7, 11, 15, …
▶️ Answer/Explanation
Step 1: Find the common difference:
\( 7 – 3 = 4 \), \( 11 – 7 = 4 \), \( 15 – 11 = 4 \)
So, common difference \( a = 4 \)
Step 2: Use the form \( an + b \)
Let’s test \( 4n \):
When \( n = 1 \), \( 4n = 4 \), but the first term is 3
So, \( b = 3 – 4 = -1 \)
Step 3: Final formula:
nth term = \( 4n – 1 \)
(b) Quadratic Sequences
A quadratic sequence has a constant second difference.
The nth term has the form: \( \text{nth term} = an^2 + bn + c \)
Example:
Find the nth term of the sequence: 2, 6, 12, 20, 30, …
▶️ Answer/Explanation
First differences:
\( 6 – 2 = 4 \), \( 12 – 6 = 6 \), \( 20 – 12 = 8 \), \( 30 – 20 = 10 \)
Second differences:
\( 6 – 4 = 2 \), \( 8 – 6 = 2 \), \( 10 – 8 = 2 \) → Constant → Quadratic
Assume form \( an^2 + bn + c \)
Use first 3 terms to form equations:
- When \( n = 1 \): \( a(1)^2 + b(1) + c = 2 \) → \( a + b + c = 2 \)
- When \( n = 2 \): \( 4a + 2b + c = 6 \)
- When \( n = 3 \): \( 9a + 3b + c = 12 \)
Solve the system:
- Subtract eq1 from eq2: \( (4a + 2b + c) – (a + b + c) = 6 – 2 \) → \( 3a + b = 4 \)
- Subtract eq2 from eq3: \( (9a + 3b + c) – (4a + 2b + c) = 12 – 6 \) → \( 5a + b = 6 \)
- Subtract: \( (5a + b) – (3a + b) = 6 – 4 \) → \( 2a = 2 \) → \( a = 1 \)
- Substitute \( a = 1 \) in \( 3a + b = 4 \) → \( 3 + b = 4 \) → \( b = 1 \)
- Substitute \( a = 1, b = 1 \) into \( a + b + c = 2 \) → \( 1 + 1 + c = 2 \) → \( c = 0 \)
Step 5: Final formula:
nth term = \( n^2 + n \)
Example:
Find the nth term of the sequence: 2, 5, 10, 17, …
▶️ Answer/Explanation
Step 1: Find the First Differences
\( 5 – 2 = 3 \)
\( 10 – 5 = 5 \)
\( 17 – 10 = 7 \)
Step 2: Find the Second Differences
\( 5 – 3 = 2 \)
\( 7 – 5 = 2 \) → Constant second difference → It’s a quadratic sequence.
Use the quadratic form:
\( \text{nth term} = an^2 + bn + c \)
Set up equations using the first three terms:
- When \( n = 1 \): \( a(1)^2 + b(1) + c = 2 \) → \( a + b + c = 2 \)
- When \( n = 2 \): \( 4a + 2b + c = 5 \)
- When \( n = 3 \): \( 9a + 3b + c = 10 \)
Solve the system of equations:
- Equation 1: \( a + b + c = 2 \)
- Equation 2: \( 4a + 2b + c = 5 \)
- Equation 3: \( 9a + 3b + c = 10 \)
Subtract Equation 1 from Equation 2:
\( (4a + 2b + c) – (a + b + c) = 5 – 2 \)
→ \( 3a + b = 3 \) → (Equation A)
Subtract Equation 2 from Equation 3:
\( (9a + 3b + c) – (4a + 2b + c) = 10 – 5 \)
→ \( 5a + b = 5 \) → (Equation B)
Subtract Equation A from Equation B:
\( (5a + b) – (3a + b) = 5 – 3 \)
→ \( 2a = 2 \) → \( a = 1 \)
Substitute \( a = 1 \) into Equation A:
\( 3(1) + b = 3 \) → \( b = 0 \)
Substitute \( a = 1, b = 0 \) into Equation 1:
\( 1 + 0 + c = 2 \) → \( c = 1 \)
Final nth term:
\( \text{nth term} = n^2 + 1 \)
(c) Cubic Sequences
A cubic sequence has a constant third difference.
Its nth term has the form: \( \text{nth term} = an^3 + bn^2 + cn + d \)
Example:
Find the nth term of the sequence: 1, 8, 27, 64, 125, …
▶️ Answer/Explanation
This is a well-known sequence:
\( 1 = 1^3 \), \( 8 = 2^3 \), \( 27 = 3^3 \), \( 64 = 4^3 \), \( 125 = 5^3 \)
nth term = \( n^3 \)