Sequences: including quadratic sequences (Higher tier only)

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Overview

A sequence is an ordered list of terms. In this topic, you move from spotting a pattern to writing a rule that calculates any term directly. That makes a question about the hundredth term a short substitution rather than a hundred separate calculations. The key Higher-tier skill is finding the nth term of a quadratic sequence, where the first differences change but the second differences are constant and non-zero.

AQA GCSE Mathematics 8300 places sequences in 3.2.4, A23–A25. Generating terms and finding linear nth terms provide the foundations. Recognising and using quadratic sequences appears in additional Foundation content; deducing a quadratic nth-term expression is Higher-only. Do not confuse these requirements. 1

Candidates may be asked to continue a sequence, describe its rule, derive an nth term, interpret a growing diagram, or decide whether a number occurs. These questions link naturally to substitution, expanding brackets, factorising and solving quadratic equations. Your goal is to show a coherent method, obtain a correct expression and interpret the answer. This guide builds that method, then tests it through examples, recall prompts and a narrated revision lesson.

Key Concepts

1. Position is not the term value

Write positions above the terms: n = 1, 2, 3, 4, … . For 5, 8, 11, 14, the third term is 11, not 3. The notation Tₙ means the value at position n. Unless a question explicitly uses a different starting index, GCSE sequences begin at n = 1.

A term-to-term rule describes how to obtain the next term from previous terms, such as ‘add 3’. A position-to-term rule uses n directly: Tₙ = 3n + 2. The second rule is more useful for a distant term: T₁₀₀ = 302. For a recursive rule, the starting term or terms must also be known.

Cover and recall: explain the difference between ‘the tenth term’ and ‘a term equal to ten’. Your explanation should distinguish a position from a value.

2. Recognise the family before choosing a method

An arithmetic, or linear, sequence has a constant first difference. Square numbers 1, 4, 9, 16, … are quadratic. Cube numbers 1, 8, 27, 64, … do not have constant second differences. Triangular numbers 1, 3, 6, 10, … are quadratic with a fractional leading coefficient.

A geometric sequence multiplies by a fixed ratio, such as 3, 6, 12, 24, … . A Fibonacci-type sequence adds the previous two terms, such as 2, 5, 7, 12, 19, … . Do not force every unfamiliar pattern into a quadratic rule. Other recursive rules will be specified when needed. 1

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3. First differences and second differences

Subtract each term from the next, consistently moving left to right. For 2, 7, 14, 23, 34, the first differences are 5, 7, 9, 11. Subtract neighbouring first differences to obtain 2, 2, 2. The constant, non-zero second difference identifies the intended quadratic pattern.

Keep negative signs: 10, 7, 2, −5 has first differences −3, −5, −7 and second differences −2, −2. A quadratic sequence need not increase. If second differences are zero throughout, the sequence is linear or constant, not genuinely quadratic.

A finite list never determines one unrestricted infinite rule uniquely. In GCSE questions, use the stated sequence type or the intended simple structure. Two matching substitutions check your arithmetic; they are not a proof about every imaginable continuation.

4. Why halving gives the quadratic coefficient

A quadratic nth term has the form Tₙ = an² + bn + c, with a ≠ 0. The square numbers have first differences 3, 5, 7, 9 and second differences 2. Therefore, multiplying square numbers by a makes the second difference 2a. Adding bn + c does not change it: a linear sequence has zero second difference.

For an algebraic explanation, Tₙ₊₁ − Tₙ = 2an + a + b. Increasing n by one changes this first difference by 2a. Hence, if the second difference is D, a = D/2. D = 6 gives a = 3, not 6. D = −4 gives a = −2. D = 1 gives a = 1/2.

Why does this work? Explain why changing c shifts every term equally but leaves both difference rows unchanged.

5. Remove the quadratic part, then rebuild

Use Differences, Halve, Subtract, Rebuild, Check. For 2, 7, 14, 23, 34, the second difference is 2, so a = 1. Subtract n², meaning subtract 1, 4, 9, 16, 25 from the corresponding original terms. The remainder is 1, 3, 5, 7, 9.

That remainder increases by 2, so begin with 2n. At n = 1, 2n gives 2 but the remainder is 1; subtract 1. Thus the remainder is 2n − 1, and the complete rule is Tₙ = n² + 2n − 1.

Check n = 1: 1 + 2 − 1 = 2. Check n = 3: 9 + 6 − 1 = 14. Candidates should subtract an² from the original sequence, not from the first differences and not in the opposite order.

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6. Find b and c without guessing

After subtracting an², the remainder is linear. Its common difference is b. If its first term is r₁, then c = r₁ − b, since b + c = r₁ at position one. In the original sequence, T₁ = a + b + c. Therefore, c is usually not the first term.

An alternative uses the first original difference d₁ = T₂ − T₁. Since d₁ = 3a + b, calculate b = d₁ − 3a and then c = T₁ − a − b. Another valid route substitutes three known positions into an² + bn + c and solves the resulting equations. Use the method you can execute reliably; AQA accepts valid alternatives according to the question-specific scheme. 5

7. Negative and fractional coefficients

For 10, 7, 2, −5, a = −1. Subtract −n²: 10 − (−1), 7 − (−4), 2 − (−9), −5 − (−16). Each remainder is 11. The rule is Tₙ = −n² + 11. There is no linear term, so b = 0.

For triangular numbers, D = 1 and a = 1/2. Subtracting (1/2)n² leaves 1/2, 1, 3/2, 2, …, which is (1/2)n. Thus Tₙ = (n² + n)/2. Keep fractions exact. A fractional coefficient does not imply fractional terms: n and n + 1 are consecutive integers, so their product is even.

8. Generate terms, test membership and justify conclusions

To find a term, substitute the requested position. For Tₙ = n² + 2n − 1, T₁₀ = 100 + 20 − 1 = 119. Follow the order of operations: square n before multiplying by its coefficient.

To test whether 47 occurs, solve n² + 2n − 1 = 47. Rearranging gives n² + 2n − 48 = 0, so (n + 8)(n − 6) = 0. The roots are −8 and 6. Only n = 6 is a positive integer, so 47 is the sixth term.

For 50, rearrangement gives (n + 1)² = 52. Since 52 is not a perfect square, neither root is an integer. Therefore, 50 is not a term. Credit for reasoning depends on a mathematical justification, not just ‘it is not in the first five terms’. Reject negative, zero or fractional positions unless the question explicitly defines a different index domain.

Mathematical/Scientific Relationships

Formula-sheet status: these labels refer to AQA’s published 2027 Higher insert. AQA confirms that a sheet is supplied, but sequence methods are not on it. 3 4 ‘Must memorise’ below means know the relationship or be able to derive it fluently, not that AQA demands rote recall of every rearrangement.

Must memorise: arithmetic nth term Tₙ = u + (n − 1)d = dn + (u − d), where u is the first term and d the common difference. Use it for a linear sequence or the linear remainder.

Must memorise: quadratic form Tₙ = an² + bn + c; first difference ΔTₙ = Tₙ₊₁ − Tₙ; second difference Δ²Tₙ = Tₙ₊₂ − 2Tₙ₊₁ + Tₙ = 2a; therefore a = D/2. For consecutive positions starting at one, b = (T₂ − T₁) − 3a and c = T₁ − a − b. The remainder is Rₙ = Tₙ − an² = bn + c.

Must memorise: standard square, cube and triangular patterns: Tₙ = n², Tₙ = n³ and Tₙ = n(n + 1)/2. A geometric sequence with first term u and ratio r has Tₙ = urⁿ⁻¹. A Fibonacci-type rule is Tₙ₊₂ = Tₙ₊₁ + Tₙ, together with two starting terms. These are supporting recognition tools; the focus here is quadratic nth terms.

Given on formula sheet: for An² + Bn + C = 0, A ≠ 0, n = (−B ± √(B² − 4AC))/(2A). Use this quadratic equation formula for a membership equation if factorising is unsuitable. Here A, B and C are the coefficients after rearranging the equation, not automatically the original sequence coefficients. Keep exact values until deciding whether a root is an integer.

Graph and Data Skills

Plot position n horizontally and term value Tₙ vertically. For n² + 2n − 1, the first five points are (1, 2), (2, 7), (3, 14), (4, 23), (5, 34). Use equally spaced scales and label both axes. The sequence contains discrete points at positive integer positions. A smooth parabola illustrates the related function but does not add fractional-position terms.

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The widening vertical gaps match increasing first differences. Positive a gives an upward-opening underlying parabola; negative a gives a downward-opening one. However, positive a does not guarantee every initial term increases, because b also affects the first differences. Do not replace an exact membership argument with a rough graph reading when an exact conclusion is required.

Practical Applications

Growing tile patterns connect diagrams to algebra. If stage n contains an n-by-n square plus 2n − 1 extra tiles, its total is n² + 2n − 1. Identify non-overlapping parts so that shared corner tiles are not counted twice. A written explanation linking each part to its expression supports a convincing ‘show that’ response.

Triangular arrangements of seats or counters give n(n + 1)/2. Two identical triangular arrangements can be paired to form an n-by-(n + 1) rectangle. This explains the division by two and links the sequence to area and expanding brackets. Such models apply only while the stated construction rule continues.

There is no required practical for this mathematics topic. Positions and tile counts are dimensionless. If terms represent lengths or costs, retain the question’s units; never attach cm² merely because n² appears. Where area conversions arise in a context, 1 cm² = 100 mm², not 10 mm².

Turning the Method into Marks

AQA June 2023 Higher Paper 2, question 19, provides a real four-mark example. Its scheme gives M1 for second difference 10 or a = 5; M1dep for removing 5n² or establishing b = 2; M1dep for a valid substitution to obtain c; and A1 for 5n² + 2n − 4. This is evidence for showing your intermediate reasoning, not a universal scheme. 5

All questions written for this guide have indicative teaching mark schemes, not official AQA allocations. On an actual paper, follow instructions such as ‘show your working’. Do not assume that a separate mark is always awarded for checking, or that follow-through is automatic.

Listen, Retrieve and Revisit

The ten-minute female-voiced lesson follows introduction, core concepts, exam technique, quick-fire recall and summary. After listening, cover the guide and reproduce the five-step method. Attempt one question now, another tomorrow and a mixed question next week. Before revealing an answer, explain why your method applies. This turns recognition into usable exam knowledge.

References