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Algebra - Sequences and General Terms

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sequence is a list of numbers in a specific order, where each number is called a 'term'.

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The position of a term is denoted by 'n' (n=1 for the 1st term, n=2 for the 2nd term, etc.).

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Arithmetic (Linear) Sequences: Sequences where the difference between consecutive terms is constant.

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Quadratic Sequences: Sequences where the first differences change, but the second differences are constant.

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Geometric Sequences: Sequences where each term is found by multiplying the previous term by a constant ratio.

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Term-to-term rule: A rule that defines how to get from one term to the next.

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Position-to-term rule (nth term): A formula that allows you to calculate any term in the sequence using its position 'n'.

📐Formulae

Linear nth term: un=dn+cu_n = dn + c (where dd is the common difference)

Alternative Linear formula: un=a+(n−1)du_n = a + (n-1)d (where aa is the first term)

Quadratic nth term: un=an2+bn+cu_n = an^2 + bn + c

Relationship for Quadratic: Second difference =2a= 2a

Geometric nth term: un=a⋅rn−1u_n = a \cdot r^{n-1} (where aa is the first term and rr is the common ratio)

Square numbers: 1,4,9,16...→n21, 4, 9, 16... \rightarrow n^2

Cube numbers: 1,8,27,64...→n31, 8, 27, 64... \rightarrow n^3

💡Examples

Problem 1:

Find the nthn^{th} term formula for the sequence: 5, 8, 11, 14, 17...

Solution:

un=3n+2u_n = 3n + 2

Explanation:

  1. Find the first difference: 8−5=3,11−8=38-5=3, 11-8=3. The difference is constant (3), so it is a linear sequence. 2. The formula starts with 3n3n. 3. To find the constant 'c', compare 3n3n to the first term: when n=1,3(1)=3n=1, 3(1)=3. We need 5, so we add 2. Formula: 3n+23n+2.

Problem 2:

Find the nthn^{th} term for the quadratic sequence: 4, 7, 12, 19, 28...

Solution:

un=n2+3u_n = n^2 + 3

Explanation:

  1. First differences: 3, 5, 7, 9. 2. Second differences: 2, 2, 2. 3. Since the second difference is 2, 2a=22a = 2, so a=1a = 1. The formula involves n2n^2. 4. Subtract n2n^2 values (1, 4, 9, 16) from the original terms: (4−1=3,7−4=3,12−9=3)(4-1=3, 7-4=3, 12-9=3). The difference is always 3. Therefore, the formula is n2+3n^2 + 3.

Problem 3:

The nthn^{th} term of a sequence is 5n−25n - 2. Find the 50th50^{th} term.

Solution:

248

Explanation:

Substitute n=50n = 50 into the general term formula: u50=5(50)−2=250−2=248u_{50} = 5(50) - 2 = 250 - 2 = 248.

Problem 4:

Identify the nthn^{th} term of the geometric sequence: 2, 6, 18, 54...

Solution:

un=2×3n−1u_n = 2 \times 3^{n-1}

Explanation:

  1. Find the common ratio: 6/2=36/2 = 3 and 18/6=318/6 = 3. 2. The first term a=2a = 2 and ratio r=3r = 3. 3. Using the geometric formula un=arn−1u_n = ar^{n-1}, we get 2×3n−12 \times 3^{n-1}.