HookWhy the March 2020 case graph terrified epidemiologists
In late February 2020 the UK had only a few dozen confirmed cases of COVID-19. By 23 March, the day lockdown was announced, the official count had passed \(6{,}500\), with cases roughly doubling every three to four days. Doubling is a sequence: \(1,\ 2,\ 4,\ 8,\ 16,\ 32,\dots\) — a geometric progression with common ratio \(2\). What alarmed epidemiologists was not the size of any single day's figure but the rule generating them. Add the same amount each day and you walk up a gentle ramp; multiply by the same amount each day and you climb a cliff. Seven doublings — about three weeks at that rate — turn one case into \(128\), and ten doublings turn it into more than a thousand.
A sequence is an ordered list of numbers built by a rule, and the whole of A2.4 is about reading that rule off a list and using it to leap ahead. You will generate terms from a term-to-term rule, a position-to-term rule and from growing patterns of dots and tiles; recognise the named families on sight — square, cube and triangular numbers, arithmetic progressions, Fibonacci-type sequences, quadratic sequences and geometric progressions; and master the two Higher headline skills: deducing a formula for the \(n\)th term of a linear and a quadratic sequence, and working with geometric progressions whose common ratio is a fraction or even a surd such as \(\sqrt{2}\). It is examined on both the calculator and non-calculator papers, and the \(n\)th-term skill is among the most reliable sources of method marks in the whole Higher course.
ModelTwo ways to describe a sequence, and generating its terms
There are exactly two ways to pin down a sequence, and telling them apart is half the battle. A term-to-term rule says how to get from one term to the next: 'start at \(4\) and add \(3\)' generates \(4,\ 7,\ 10,\ 13,\ 16\). It is easy to apply but has a weakness — to reach the \(50\)th term you must grind through all \(49\) before it, because each term leans on the one in front.
A position-to-term rule removes that limit by giving a formula in the position number \(n\): the rule \(3n+1\) lets you drop any position straight in. Want the \(50\)th term? Substitute \(n=50\) to get \(3\times 50+1=151\), with no list. The position \(n\) is always a whole counting number — \(1\) for the first term, \(2\) for the second — so you never substitute \(n=0\) or a fraction. This position rule is what the exam calls the \(n\)th term, and it is the single most useful idea in the section.
The practical skill is generating terms cleanly from whichever rule you are handed, and reading a growing pattern of matchsticks or tiles as a sequence by counting the first few arrangements. The classic slip is muddling the two kinds of rule — feeding a term back in when the formula actually wants the position number — so decide which rule you have before you write anything.
A sequence has first term \(5\) and the term-to-term rule 'multiply by \(2\), then subtract \(3\)'. Write the first five terms. Start at \(5\). Next: \(5\times 2-3=7\). Next: \(7\times 2-3=11\). Next: \(11\times 2-3=19\). Next: \(19\times 2-3=35\). So the sequence is \(5,\ 7,\ 11,\ 19,\ 35\). You could not have jumped to the fifth term — each needed the one before, exactly the limitation a position rule removes. Compare the position rule \(4n-1\): substituting \(n=1,\ 2,\ 3,\ 4\) gives \(3,\ 7,\ 11,\ 15\) at once, and its \(100\)th term is instant, \(4\times 100-1=399\).
MechanismThe named families — identify by differences and ratios
AQA expects instant recognition of several families, because questions often hinge on naming one. Square numbers are \(1,\ 4,\ 9,\ 16,\ 25\) (that is \(1^2,\ 2^2,\ 3^2,\dots\)); cube numbers are \(1,\ 8,\ 27,\ 64,\ 125\) (\(1^3,\ 2^3,\ 3^3,\dots\)); triangular numbers are \(1,\ 3,\ 6,\ 10,\ 15\), the dots in a growing triangle, adding \(2\), then \(3\), then \(4\).
An arithmetic (linear) sequence climbs by a fixed common difference: \(5,\ 8,\ 11,\ 14\) adds \(3\). A geometric sequence multiplies by a fixed common ratio: \(3,\ 6,\ 12,\ 24\) doubles. A Fibonacci-type sequence adds the previous two terms: \(1,\ 1,\ 2,\ 3,\ 5,\ 8\), or any starting pair such as \(4,\ 7,\ 11,\ 18\). And a quadratic sequence — new prominence at Higher — has a constant second difference, like the square numbers.
The skill that ties them together is a quick diagnostic. Work out the first differences: if they are constant, the sequence is arithmetic. If not, check the ratios (each term divided by the one before): a constant ratio means geometric. If neither, find the second differences: a constant one means quadratic. And if each term is the sum of the two before, it is Fibonacci-type. That thirty-second test tells you what you are looking at before you try to extend or generalise it.
Classify \(2,\ 6,\ 12,\ 20,\ 30\). First differences: \(4,\ 6,\ 8,\ 10\) — not constant, so not arithmetic. Ratios: \(6\div 2=3\) but \(12\div 6=2\) — not constant, so not geometric. Second differences: \(6-4=2,\ 8-6=2,\ 10-8=2\) — constant, so it is a quadratic sequence. To extend it, the next first difference is \(10+2=12\), so the next term is \(30+12=42\). (Its formula is \(n^2+n\), which a later block shows how to find.)
MechanismThe nth term of a linear sequence
Finding the \(n\)th-term rule of an arithmetic sequence is a fixed two-step method. Step one: the common difference \(d\) becomes the coefficient of \(n\). Step two: find the adjustment by comparing \(dn\) against the real first term. For \(5,\ 8,\ 11,\ 14\), the common difference is \(3\), so start with \(3n\); but \(3n\) gives \(3,\ 6,\ 9,\ 12\), each \(2\) short of the sequence, so the rule is \(3n+2\).
Why does the difference become the coefficient? A linear sequence adds the same amount every step, and \(3n\) is precisely the three-times table, which also adds \(3\) every step. Your sequence marches in lockstep with that table, shifted up or down by a constant — and that constant shift is what you add. A descending sequence works identically with a negative coefficient: \(20,\ 17,\ 14,\ 11\) has \(d=-3\), so start \(-3n\); since \(-3n\) gives \(-3,\ -6,\ -9\) while the sequence is \(20,\ 17,\ 14\), the adjustment is \(+23\), giving \(-3n+23\).
The formula then answers 'is this number a term?' Set the rule equal to the target and solve for \(n\); if \(n\) is a whole number the value is in the sequence, and if not, it is not — and saying so explicitly is the mark.
Find the \(n\)th term of \(7,\ 12,\ 17,\ 22\), then decide whether \(158\) is a term. The common difference is \(5\), so start with \(5n\). The five-times table \(5,\ 10,\ 15,\ 20\) is \(2\) short of the sequence, so add \(2\): the \(n\)th term is \(5n+2\). Check: \(n=1\) gives \(7\) ✓, \(n=4\) gives \(22\) ✓. Now test \(158\): solve \(5n+2=158\), so \(5n=156\), \(n=31.2\). That is not a whole number, so \(158\) is not in the sequence. By contrast \(157\) gives \(5n=155\), \(n=31\) — a whole number, so \(157\) is the \(31\)st term.
ModelThe nth term of a quadratic sequence — the Higher headline
Deducing the \(n\)th term of a quadratic sequence is the skill that separates Higher from Foundation, and it has a clean recipe. Every quadratic sequence has an \(n\)th term of the form \(an^2+bn+c\). The key fact is that the second difference equals \(2a\), so \(a\) is half the second difference. Once you have \(a\), subtract \(an^2\) from the original sequence: what remains is always a linear sequence, whose \(n\)th term \(bn+c\) you find by the two-step method from the previous block. Put the pieces together and you have the full quadratic rule.
The logic is worth seeing. Differencing a linear rule once gives a constant; differencing a quadratic rule twice gives a constant, and that constant is exactly \(2a\). So the size of the second difference tells you the \(n^2\) coefficient immediately, and stripping the \(n^2\) part away leaves a simpler problem you already know how to solve. This 'peel off the square, then finish linearly' structure is the heart of every quadratic-sequence question.
Find the \(n\)th term of \(3,\ 8,\ 15,\ 24,\ 35\). First differences: \(5,\ 7,\ 9,\ 11\). Second difference: \(2\). So \(2a=2\), giving \(a=1\) and an \(n^2\) part of \(1\times n^2=n^2\), namely \(1,\ 4,\ 9,\ 16,\ 25\). Subtract this from the sequence: \(3-1=2,\ 8-4=4,\ 15-9=6,\ 24-16=8,\ 35-25=10\), leaving \(2,\ 4,\ 6,\ 8,\ 10\). That linear remainder has common difference \(2\) and passes through \(2\) at \(n=1\), so it is \(2n\). Combine: the \(n\)th term is \(n^2+2n\). Check \(n=5\): \(25+10=35\) ✓, and \(n=3\): \(9+6=15\) ✓. Always verify on two terms before trusting the formula.
DataGeometric progressions — rational and surd ratios
A geometric progression multiplies by a fixed common ratio \(r\) each step, and because multiplying compounds, it grows or shrinks far faster than any arithmetic sequence — the shape of the March 2020 case curve. Find the ratio by dividing any term by the one before: for \(3,\ 6,\ 12,\ 24\) the ratio is \(6\div 3=2\), confirmed by \(12\div 6=2\); continue by multiplying, so the next term is \(24\times 2=48\). With first term \(a\) and ratio \(r\), the \(n\)th term is \(a\,r^{\,n-1}\), which lets you reach a distant term without listing.
Higher widens the ratios you must handle. A ratio can be a fraction less than \(1\), giving a decaying geometric sequence such as \(80,\ 40,\ 20,\ 10\) with \(r=\tfrac{1}{2}\) — still geometric, just shrinking. More distinctively, the ratio can be a surd: a sequence such as \(2,\ 2\sqrt{3},\ 6,\ 6\sqrt{3},\ 18\) multiplies by \(\sqrt{3}\) each time, because \(2\sqrt{3}\times\sqrt{3}=2\times 3=6\). Surd ratios are where Higher tests your surd arithmetic and your sequence work at once, so the step \(\sqrt{3}\times\sqrt{3}=3\) must be automatic.
The Fibonacci sequence is not geometric — it adds rather than multiplies — but it hides a geometric ghost: divide consecutive Fibonacci terms and the answers home in on the golden ratio, roughly \(1.618\). You are not asked to prove that, but distinguishing 'multiply by a fixed ratio' (geometric) from 'add the previous two' (Fibonacci-type) is exactly the naming the exam rewards.
A geometric progression has first term \(5\) and common ratio \(\sqrt{2}\). Find the first four terms and the \(7\)th term. Multiply repeatedly: \(5,\ 5\sqrt{2},\ 5\times(\sqrt{2})^2=10,\ 10\sqrt{2}\). For the \(7\)th term use \(a\,r^{\,n-1}=5\times(\sqrt{2})^{6}\). Now \((\sqrt{2})^{6}=\big((\sqrt{2})^2\big)^3=2^3=8\), so the \(7\)th term is \(5\times 8=40\). The trick every surd-ratio question rewards is turning \((\sqrt{2})^{6}\) into \(2^{3}\) rather than reaching for a decimal — exact from start to finish.
CaseSequences from patterns — where a picture becomes a formula
A favourite Higher question hands you a growing pattern of tiles or dots and asks for the count in the \(50\)th or \(n\)th arrangement. Trying to draw large cases is a trap; the method is to count the first few, tabulate them as a sequence, and generalise. If the counts have a constant first difference the pattern is linear and you use the \(dn+a\) method; if they have a constant second difference the pattern is quadratic and you use the half-second-difference recipe.
Reading why a pattern grows as it does bridges the picture and the formula. A single row of squares made from matchsticks needs \(4\) matches for one square, \(7\) for two, \(10\) for three: each extra square adds only \(3\) matches because it borrows a side from its neighbour. That structural '\(3\) more each time' is the \(3n\) in the rule, and the leftover match is the adjustment. When a pattern grows in two directions at once — a solid square of dots, say — the count grows quadratically, and the second-difference method delivers the \(n^2\) formula.
A pattern of dots grows: pattern \(1\) has \(2\) dots, pattern \(2\) has \(6\), pattern \(3\) has \(12\), pattern \(4\) has \(20\). Find a formula for pattern \(n\). Tabulate \(2,\ 6,\ 12,\ 20\): first differences \(4,\ 6,\ 8\) are not constant, but the second differences \(2,\ 2\) are, so the pattern is quadratic. Half the second difference gives \(a=1\), so the \(n^2\) part is \(1,\ 4,\ 9,\ 16\); subtracting leaves \(2-1=1,\ 6-4=2,\ 12-9=3,\ 20-16=4\), the linear sequence \(n\). So pattern \(n\) has \(n^2+n\) dots. Pattern \(50\) therefore has \(50^2+50=2550\) dots — found without drawing a single one.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Sequences appear on both papers, and the \(n\)th-term question is one of the most predictable mark-earners in the course, so drill the methods until they are automatic. For a linear sequence, write the common difference as the coefficient of \(n\) and show the \(dn\) line before you fix the constant — that line is a method mark even if the adjustment is wrong. For a quadratic sequence, show the second difference, halve it for \(a\), write out \(an^2\), subtract it, then solve the linear remainder; examiners award marks for each stage, so a bare final formula throws marks away.
Always verify a rule on two known terms (\(n=1\) and \(n=2\)) before trusting it for a distant term, and keep negatives intact on descending sequences — a common difference of \(-3\) makes the rule start \(-3n\), and dropping the sign is the commonest slip. Read the command precisely: 'write the next two terms' wants a term-to-term step; 'find the \(n\)th term' wants the position formula; 'is \(158\) a term?' wants you to solve \(dn+a=158\) and state whether \(n\) is a whole number. On geometric progressions keep surds exact — turn \((\sqrt{2})^{6}\) into \(2^{3}\) rather than a rounded decimal — and name families explicitly, since 'geometric, ratio \(\sqrt{3}\)' or 'quadratic, second difference \(2\)' is often worth a mark on its own.