Arithmetic Sequences, Quadratic nth Term & Geometric Sequences
A sequence is an ordered list of numbers following a mathematical rule that connects each term to its position n (the position-to-term or nth term rule) or to the previous term (term-to-term / iteration rule). Linear (arithmetic) sequences have a constant first difference d between consecutive terms, giving the nth term formula dn + c, whereas quadratic sequences on the Higher Tier have a constant second difference 2a, giving the formula an² + bn + c. Examiners frequently test finding linear and quadratic nth terms, proving algebraically whether a given number is a term in a sequence, and working with geometric progressions (with surd or fractional common ratios) and Fibonacci-style sequences.
- Linear (Arithmetic) Sequences & nth Term
- In an arithmetic sequence, the difference (d) between consecutive terms is constant. The nth term formula is dn + c, where d is the common difference and c is the 0th term (found by subtracting d from the 1st term). If a sequence is decreasing (e.g., 20, 17, 14, 11.), d is negative (-3), so the nth term is -3n + 23 (or 23 - 3n).
- Testing Membership in a Sequence
- To determine if a value V belongs to a sequence (or to find which term is the first to exceed V), form the equation (or inequality) nth term = V and solve for n. Because position n can only take positive integer values (n = 1, 2, 3.), V is a term in the sequence if and only if n is a positive integer.
- Quadratic Sequences (an² + bn + c) (Higher Tier)
- A quadratic sequence has a changing first difference, but a constant second difference. For nth term U_n = an² + bn + c: 1. The second difference is always equal to 2a, so a = (Second Difference) ÷ 2. 2. Subtract the sequence an² from the given sequence U_n to leave a simple linear sequence (bn + c). 3. Find the linear nth term bn + c and combine into an² + bn + c.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Linear (Arithmetic) Sequences & nth Term
- In an arithmetic sequence, the difference (d) between consecutive terms is constant. The nth term formula is dn + c, where d is the common difference and c is the 0th term (found by subtracting d from the 1st term). If a sequence is decreasing (e.g., 20, 17, 14, 11.), d is negative (-3), so the nth term is -3n + 23 (or 23 - 3n).
- Testing Membership in a Sequence
- To determine if a value V belongs to a sequence (or to find which term is the first to exceed V), form the equation (or inequality) nth term = V and solve for n. Because position n can only take positive integer values (n = 1, 2, 3.), V is a term in the sequence if and only if n is a positive integer.
- Quadratic Sequences (an² + bn + c) (Higher Tier)
- A quadratic sequence has a changing first difference, but a constant second difference. For nth term U_n = an² + bn + c: 1. The second difference is always equal to 2a, so a = (Second Difference) ÷ 2. 2. Subtract the sequence an² from the given sequence U_n to leave a simple linear sequence (bn + c). 3. Find the linear nth term bn + c and combine into an² + bn + c.
- Geometric Sequences (Common Ratio r)
- A sequence where each term is found by multiplying the previous term by a non-zero constant called the common ratio (r). On the Higher Tier, r can be a fraction, negative number, or a surd (e.g., √2, 2, 2√2, 4. has common ratio r = √2). If consecutive terms are algebraic expressions (e.g., x - 1, 2x, 5x + 3), equate the common ratios: (2nd term) / (1st term) = (3rd term) / (2nd term) and cross-multiply to solve for x.
- Fibonacci-Type & Recurrence Sequences
- In a Fibonacci-type sequence, each term is the sum of the two previous terms: U_(n+2) = U_(n+1) + U_n. When given algebraic first and second terms (e.g., a and b), generate the subsequent terms algebraically by adding consecutive pairs: 1st = a, 2nd = b, 3rd = a + b, 4th = a + 2b, 5th = 2a + 3b, 6th = 3a + 5b.
- Triangular, Square & Cube Number Sequences
- Special sequences to recognise on sight: 1. Square numbers (n²): 1, 4, 9, 16, 25, 36. 2. Cube numbers (n³): 1, 8, 27, 64, 125. 3. Triangular numbers (½n(n + 1)): 1, 3, 6, 10, 15, 21. (first differences are +2, +3, +4, +5.).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Find the nth term of the decreasing linear sequence: 25, 21, 17, 13, 9.
In a quadratic sequence an² + bn + c, if the second difference is +6, what is the value of a?
How do you find the next two terms of the geometric sequence: 3√2, 6, 6√2, .?
The first two terms of a Fibonacci-style sequence are x and y. Write down the 3rd, 4th, and 5th terms in terms of x and y.
A sequence is defined by the iteration rule U_(n+1) = 2U_n - 3, with U₁ = 5. Find U₂ and U₃.
What is the formula for the nth triangular number (1, 3, 6, 10, 15.)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Here are the first five terms of a quadratic sequence: 4, 11, 22, 37, 56. Find an expression, in terms of n, for the nth term of this sequence. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Find the first differences between consecutive terms: +7, +11, +15, +19; then find the second differences: +4, +4, +4 [1 mark]. 2. Halve the constant second difference (+4) to find the coefficient of n²: a = 4 ÷ 2 = 2, so the quadratic term is 2n² [1 mark]. 3. Subtract 2n² (which for n = 1, 2, 3, 4 is 2, 8, 18, 32) from the original sequence (4, 11, 22, 37): Remainder sequence = 2, 3, 4, 5 [1 mark]. 4. The linear nth term of 2, 3, 4, 5 is n + 1; combining gives the final nth term = 2n² + n + 1 [1 mark].
- Q2
The nth term of an arithmetic sequence is 7n - 5. (a) Explain why 298 cannot be a term in this sequence. (b) Find the value and position of the first term in the sequence that is greater than 400. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Set the nth term equal to 298 and solve for n: 7n - 5 = 298 -> 7n = 303 -> n = 303 ÷ 7 = 43.285. [1 mark]. 2. State conclusion: Since n must be a whole number (positive integer) and 43.285. is not an integer, 298 is not a term in the sequence [1 mark]. 3. (b) Set up the inequality 7n - 5 > 400 -> 7n > 405 -> n > 405 ÷ 7 = 57.857. [1 mark]. 4. The smallest integer greater than 57.857. is n = 58 ( the 58th term), which has value 7(58) - 5 = 406 - 5 = 401 [1 mark].
- Q3
The first three terms of a geometric sequence with positive terms are (x - 3), (x + 1), and (2x + 8). Work out the numerical value of the 4th term of this sequence. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. In a geometric sequence, the ratio of consecutive terms is constant: (2nd term) / (1st term) = (3rd term) / (2nd term), so (x + 1) / (x - 3) = (2x + 8) / (x + 1) [1 mark]. 2. Cross-multiply to eliminate fractions: (x + 1)(x + 1) = (2x + 8)(x - 3) [1 mark]. 3. Expand both sides and rearrange to a quadratic equation (= 0): x² + 2x + 1 = 2x² + 2x - 24 -> x² - 25 = 0 [1 mark]. 4. Factorise and solve: (x - 5)(x + 5) = 0 -> x = 5 (reject x = -5 as the terms are positive); substituting x = 5 gives the first three terms as 2, 6, and 18 with common ratio r = 6 ÷ 2 = 3 [1 mark]. 5. Multiply the 3rd term by r = 3 to find the 4th term: 18 x 3 = 54 [1 mark].
- Q4
In a Fibonacci-type sequence, each term after the first two is the sum of the two previous terms. The 1st term is a and the 2nd term is b. Given that the 3rd term is 9 and the 6th term is 39, find the values of a and b. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Write down the first six terms algebraically in terms of a and b: 1st = a, 2nd = b, 3rd = a + b, 4th = a + 2b, 5th = 2a + 3b, 6th = 3a + 5b [1 mark]. 2. Form two simultaneous equations from the 3rd and 6th terms: a + b = 9 and 3a + 5b = 39 [1 mark]. 3. Multiply the first equation by 3 (3a + 3b = 27) and subtract from the second equation: (3a + 5b) - (3a + 3b) = 39 - 27 -> 2b = 12 -> b = 6 [1 mark]. 4. Substitute b = 6 into a + b = 9: a + 6 = 9 -> a = 3 [1 mark].
Common Exam Questions (4)
How do you find the nth term of a linear (arithmetic) sequence like 7, 11, 15, 19.?
1. Find the common first difference between terms: here it increases by +4 each time, so write '4n'. 2. Find the '0th term' (what would come immediately before the 1st term by working backwards: 7 - 4 = +3). 3. Combine them: nth term = 4n + 3. Always check by substituting n = 3: 4(3) + 3 = 15 (correct).
How do you prove whether a number (like 150) is in a sequence with nth term 4n + 3?
Set the nth term formula equal to the target number (4n + 3 = 150) and solve the equation for n! Since n represents the position number of the term (1st, 2nd, 3rd.), n must be a positive whole number (integer): 1. If n is a whole number, the value IS in the sequence (at position n). 2. If n is a decimal or fraction (here 4n = 147 -> n = 36.75), state that 'Since n = 36.75 is not an integer, 150 is not in the sequence'.
What is the step-by-step method for finding the nth term of a quadratic sequence (an² + bn + c)? (Higher Tier)
Use the 'Halve the Second Difference & Subtract an²' method. 1. Work out the first differences between terms, and then the second differences (the difference of the differences). 2. Halve the constant second difference to find 'a' (since 2a = second difference). 3. Write out the first four values of an² (for n = 1, 2, 3, 4) and subtract them from the original sequence terms. 4. Find the linear nth term (bn + c) of the remaining sequence and add it to an²!
What is a geometric sequence and how do you find its common ratio r?
In a geometric sequence, you multiply by the same constant number (the common ratio, r) to get from one term to the next. To find the common ratio r, divide any term by the term immediately before it: r = (2nd term) / (1st term) = (3rd term) / (2nd term). The nth term of a geometric sequence with first term a and common ratio r is a x r^(n-1).
Need more questions on Arithmetic Sequences, Quadratic nth Term & Geometric Sequences, or want to organise your own notes?
Type a specific exam question you are stuck on, pick one of the prompts below, or paste your class notes to build a custom set of flashcards and practice questions on this page.
Related study guides
Algebra: Solving Equations & Inequalities
Algebra uses letters and symbols to represent variables, formulate equations, and solve mathematical problems systematically. For GCSE Maths, fluency with expanding brackets, factorising expressions, rearranging formulae, and solving linear, quadratic, and simultaneous equations underpins both Foundation and Higher papers.
Quadratic Equations & The Quadratic Formula
A quadratic expression takes the form ax² + bx + c (where a ≠ 0) and plots as a symmetrical parabola. In GCSE Maths, you solve quadratic equations by factorising into two brackets, using the quadratic formula, or completing the square to find the roots, y-intercept, and turning point.
Laws of Indices, Fractional & Negative Powers, and Simplifying Surds
Laws of indices provide the algebraic rules for multiplying, dividing, and raising powers with the same base, extending on the Higher Tier to negative indices (which take the reciprocal, x⁻ⁿ = 1/xⁿ) and fractional indices (where the denominator is the root and the numerator is the power, x^(m/n) = (ⁿ√x)ᵐ). Surds are irrational roots of integers (such as √2 or √5) that cannot be written as exact fractions and must be manipulated in exact form on Non-Calculator Paper 1. Examiners regularly test evaluating combined negative fractional powers of fractions, simplifying surd expressions, and rationalising two-term denominators using the difference of two squares conjugate.