Algebra
Grade 6-9Quadratic & Geometric Sequences
Beyond simple linear sequences, GCSE Maths also tests quadratic sequences, which grow by a changing amount each time, and geometric sequences, which grow by a constant multiplying factor. This lesson covers evaluating a quadratic sequence, finding the nth term of a quadratic sequence, finding the common ratio of a geometric sequence, and solving problems involving geometric sequences.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Evaluating a quadratic sequence
Substitute the position number into the nth term formula, which includes an n² term, and work out the result following the order of operations.
For the nth term n² + 3, the 1st term is 1² + 3 = 4, and the 4th term is 4² + 3 = 19.
Finding the nth term of a quadratic sequence
Find the first differences between consecutive terms, then find the second differences between those. Half the constant second difference gives the coefficient of n². Subtract this coefficient times n² from the original sequence to find the remaining linear part.
For 3, 8, 15, 24, 35, the first differences are 5, 7, 9, 11, and the second difference is 2, so the coefficient of n² is 1. Subtracting n² (giving 1, 4, 9, 16, 25) from the sequence leaves 2, 4, 6, 8, 10, which is 2n. The nth term is n² + 2n.
Finding the common ratio of a geometric sequence
Divide any term by the term before it to find the common ratio, then multiply by this ratio to find later terms.
For 2, 6, 18, 54, ..., the common ratio is 6 ÷ 2 = 3, so the next term is 54 × 3 = 162.
Solving geometric sequence problems
The nth term of a geometric sequence is a × rⁿ⁻¹, where a is the first term and r is the common ratio. If two non-consecutive terms are known, dividing one by the other gives a power of r that can be used to find r itself.
If the 2nd term of a geometric sequence is 6 and the 5th term is 162, dividing gives r³ = 162 ÷ 6 = 27, so r = 3. Substituting back, the 2nd term is a × 3 = 6, so a = 2.
Worked Examples
Three exam-style questions, fully solved.
The nth term of a sequence is n² + 3. Find the 1st and 4th terms.
Easy- 1.Substitute n = 1: 1² + 3
- 2.Substitute n = 4: 4² + 3
Answer: 1st term = 4, 4th term = 19
Find the nth term of the sequence 3, 8, 15, 24, 35, ...
Medium- 1.Find the first differences: 5, 7, 9, 11
- 2.Find the second difference: 2, so the coefficient of n² is 2 ÷ 2 = 1
- 3.Subtract n² (1, 4, 9, 16, 25) from the sequence: 2, 4, 6, 8, 10, which is 2n
Answer: n² + 2n
In a geometric sequence, the 2nd term is 6 and the 5th term is 162. Find the common ratio and the first term.
Hard- 1.Divide the 5th term by the 2nd term to find r³: 162 ÷ 6 = 27, so r = 3
- 2.Substitute into the 2nd term, a × r = 6: a × 3 = 6
Answer: r = 3, a = 2
Avoid These
The most common mistakes students make.
Making an arithmetic slip when evaluating a quadratic nth term, especially forgetting to square n before multiplying or adding the other terms.
When finding the nth term of a quadratic sequence, forgetting to halve the constant second difference to find the coefficient of n², or forgetting to subtract the n² part from the original sequence to find the remaining linear term.
Confusing a quadratic sequence, which has a constant second difference, with a geometric sequence, which has a constant ratio, and trying to apply the wrong method.
When finding the common ratio, dividing the terms in the wrong order, or getting the sign wrong for a sequence with alternating positive and negative terms.
In a geometric sequence problem using two non-consecutive terms, using the wrong power of r, forgetting that terms three positions apart give r³, not r² or r⁴.
FAQ
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