Sequences and Series

This lesson covers Sequences and Series as required by the A-Level Mathematics Pure 1 specification. A sequence is an ordered list of numbers following a rule, and a series is the sum of the terms of a sequence. This topic includes arithmetic and geometric sequences, sigma notation, the binomial expansion, and recurrence relations.

Spec Mapping — AQA 7357 Section D Sequences and series. This lesson covers arithmetic and geometric progressions, sigma notation, the binomial expansion and recurrence relations from Section D. Refer to the official AQA specification document for exact wording.

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Arithmetic Sequences

An arithmetic sequence has a constant difference between consecutive terms. This difference is called the common difference d.

Key Formulae

Formula	Description
uₙ = a + (n − 1)d	nth term
Sₙ = n/2 × (2a + (n − 1)d)	Sum of first n terms
Sₙ = n/2 × (a + l)	Sum using first and last terms

where a is the first term, d is the common difference, and l is the last term.

Example: The 5th term of an arithmetic sequence is 17 and the 12th term is 38. Find a and d.

u₅ = a + 4d = 17
u₁₂ = a + 11d = 38

Subtracting: 7d = 21 → d = 3
Substituting: a = 17 − 12 = 5

So uₙ = 5 + 3(n − 1) = 3n + 2.

Example: Find the sum of the first 20 terms when a = 4, d = 3.

S₂₀ = 20/2 × (2(4) + 19(3)) = 10 × (8 + 57) = 10 × 65 = 650

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Geometric Sequences

A geometric sequence has a constant ratio between consecutive terms. This ratio is called the common ratio r.

Key Formulae

Formula	Description
uₙ = arⁿ⁻¹	nth term
Sₙ = a(1 − rⁿ) / (1 − r)	Sum of first n terms (r ≠ 1)
S∞ = a / (1 − r)	Sum to infinity (only when

Example: The 3rd term of a geometric sequence is 12 and the 6th term is 96. Find a and r.

u₃ = ar² = 12
u₆ = ar⁵ = 96

Dividing: r³ = 96/12 = 8 → r = 2
Substituting: a × 4 = 12 → a = 3

Convergent Geometric Series

A geometric series converges (has a finite sum to infinity) if and only if |r| < 1.

Example: Find the sum to infinity of 8 + 4 + 2 + 1 + ...

Here a = 8 and r = 1/2. Since |r| < 1:

$$S∞ = 8 / (1 - 1/2) = 8 / (1/2) = 16$$S∞=8/(1−1/2)=8/(1/2)=16

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Sigma Notation

Sigma notation Σ provides a concise way to write a sum:

Σ (from r = 1 to n) of uᵣ = u₁ + u₂ + u₃ + ... + uₙ

Standard Results

Sum	Formula
Σₖ₌₁ⁿ k	n(n + 1)/2
Σₖ₌₁ⁿ k²	n(n + 1)(2n + 1)/6

Example: Evaluate Σₖ₌₁²⁰ (3k + 1).

= 3 Σₖ₌₁²⁰ k + Σₖ₌₁²⁰ 1
= 3 × 20(21)/2 + 20
= 3 × 210 + 20
= 650

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Recurrence Relations

A recurrence relation defines each term using the previous term(s).

Example: Given uₙ₊₁ = 3uₙ − 2, u₁ = 4, find u₂, u₃, u₄.

$$\begin{aligned} u_{2} = 3(4) - 2 = 10 \\ u_{3} = 3(10) - 2 = 28 \\ u_{4} = 3(28) - 2 = 82 \end{aligned}$$u2​=3(4)−2=10u3​=3(10)−2=28u4​=3(28)−2=82​

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The Binomial Expansion

The binomial expansion of (a + b)ⁿ for positive integer n is:

(a + b)ⁿ = Σ (from k = 0 to n) of ⁿCₖ × aⁿ⁻ᵏ × bᵏ

where ⁿCₖ = n! / (k!(n − k)!).

Pascal's Triangle

Pascal's triangle provides the binomial coefficients. Each entry is the sum of the two entries above it.

Row 0:           1
Row 1:         1   1
Row 2:       1   2   1
Row 3:     1   3   3   1
Row 4:   1   4   6   4   1

Example: Expand (x + 2)⁴.

= ⁴C₀ x⁴ + ⁴C₁ x³(2) + ⁴C₂ x²(4) + ⁴C₃ x(8) + ⁴C₄(16)
= x⁴ + 8x³ + 24x² + 32x + 16

Finding a Specific Term

Example: Find the coefficient of x³ in the expansion of (3 + 2x)⁷.

The general term is ⁷Cₖ × 3⁷⁻ᵏ × (2x)ᵏ. For x³, set k = 3:

$\binom{7}{3} \times 3^{4} \times 2^{3} = 35 \times 81 \times 8 = 22{,}680$(37​)×34×23=35×81×8=22,680

Approximations

Binomial expansions can approximate values when x is small.

Example: Use (1 + x)¹⁰ up to x² to approximate 1.02¹⁰.

(1 + 0.02)¹⁰ ≈ 1 + 10(0.02) + 45(0.02)² = 1 + 0.2 + 0.018 = 1.218

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Synoptic Links

This topic connects to:

• Binomial expansion for fractional/negative indices (aqa-alevel-maths-pure-2 / binomial-expansion-extended) — extends the positive-integer expansion to convergent series.
• Exponential growth and decay (aqa-alevel-maths-pure-1 / exponentials-and-logarithms) — geometric sequences are the discrete analogue of continuous exponential models.
• Probability distributions (aqa-alevel-maths-statistics / binomial-distribution) — the binomial coefficients here are exactly the probabilities in the binomial distribution.

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Summary

• Arithmetic sequences: constant difference d; use uₙ = a + (n − 1)d and Sₙ = n/2 × (2a + (n − 1)d).
• Geometric sequences: constant ratio r; use uₙ = arⁿ⁻¹ and Sₙ = a(1 − rⁿ)/(1 − r).
• A geometric series converges if |r| < 1, with S∞ = a/(1 − r).
• Sigma notation Σ is a compact way to write sums.
• Recurrence relations define terms using previous terms.
• The binomial expansion gives the terms of (a + b)ⁿ using ⁿCₖ.

Exam Tip: In questions about sequences, always identify whether the sequence is arithmetic or geometric first. For binomial expansion questions, clearly state the values of n, a, b, and k before substituting. Many marks are lost through arithmetic errors in these calculations — show every step. When asked for a specific term, write out the general term formula first.

A-Level Deep Dive: Sequences and Series

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section D — Sequences and Series covers generate sequences from a formula for the $n$nth term and from a recurrence relation; understand and use sigma notation; arithmetic and geometric sequences and series; sums to a finite term; sum to infinity of a convergent geometric series; binomial expansion of $(1 + x)^n$(1+x)n for positive integer $n$n and the extension to rational $n$n for $|x| < 1$∣x∣<1 (refer to the official specification document for exact wording). The AQA formula booklet does list the closed forms for arithmetic sums ($S_n = \tfrac{n}{2}(2a + (n - 1)d)$Sn​=2n​(2a+(n−1)d)), geometric sums ($S_n = a(1 - r^n)/(1 - r)$Sn​=a(1−rn)/(1−r)) and the geometric sum to infinity ($S_\infty = a/(1 - r)$S∞​=a/(1−r) for $|r| < 1$∣r∣<1), so candidates are not penalised for re-deriving them — but they must quote them precisely.

Synoptically, sequences and series threads through almost every other AQA Pure section: the binomial expansion is itself a finite/infinite series; integration as a Riemann sum is a limit of an arithmetic-style partition; logarithm equations $r^n = k$rn=k inside geometric sums require Section E (Exponentials and logarithms); and modelling questions in Paper 3 routinely set up GPs for compound interest, depreciation, or radioactive-decay step models. A candidate fluent in this section unlocks marks across the entire qualification.

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks):

The first three terms of an arithmetic progression are $4$4, $4 + d$4+d and $4 + 2d$4+2d, where $d > 0$d>0. The numbers $4 + 2d$4+2d, $4 + d$4+d and a third number $T$T form, in that order, the first three terms of a geometric progression with common ratio $r$r, where $|r| < 1$∣r∣<1.

(a) Show that $(4 + d)^2 = (4 + 2d) T$(4+d)2=(4+2d)T, and that $r = \dfrac{4 + d}{4 + 2d}$r=4+2d4+d​. (3)

(b) Given a clean numerical setup, find $d$d, $r$r, and the sum to infinity of the GP. (5)

Solution with mark scheme:

(a) Step 1 — apply the GP common-ratio condition.

For a GP, the ratio of consecutive terms is constant: $\dfrac{4 + d}{4 + 2d} = \dfrac{T}{4 + d} = r$4+2d4+d​=4+dT​=r.

M1 — writing the GP common-ratio relation between consecutive pairs.

Step 2 — cross-multiply the first equality.

$(4 + d)(4 + d) = (4 + 2d) T$(4+d)(4+d)=(4+2d)T, i.e. $(4 + d)^2 = (4 + 2d) T$(4+d)2=(4+2d)T.

A1 — printed result.

Step 3 — read off $r$r.

From $\dfrac{4 + d}{4 + 2d} = r$4+2d4+d​=r, we have $r = \dfrac{4 + d}{4 + 2d}$r=4+2d4+d​ as required.

A1 — second printed result.

(b) Step 1 — substitute the data given.

Substitute the prescribed value of $T$T into $(4 + d)^2 = (4 + 2d) T$(4+d)2=(4+2d)T to form a single equation in $d$d alone.

M1 — substituting and forming an equation in $d$d.

Step 2 — solve the resulting quadratic in $d$d.

Expand and rearrange to a quadratic; solve via factorisation, the quadratic formula, or completing the square. Discard any root with $d \leq 0$d≤0 since the question stipulates $d > 0$d>0.

A1 — correct $d$d value with the negative root rejected explicitly.

Step 3 — compute $r$r and $S_\infty$S∞​.

With $d$d now known, evaluate $r = (4 + d)/(4 + 2d)$r=(4+d)/(4+2d) numerically. Check $|r| < 1$∣r∣<1 — if not, the sum to infinity does not exist and the question setup is internally inconsistent. The first GP term is $4 + 2d$4+2d, so the sum to infinity is

$S_\infty = \dfrac{4 + 2d}{1 - r}$S∞​=1−r4+2d​

M1 — applying $S_\infty = \dfrac{a}{1 - r}$S∞​=1−ra​ to the GP first term $a = 4 + 2d$a=4+2d and common ratio $r$r.

A1 — correct $r$r.

A1 — correct $S_\infty$S∞​ with the convergence check $|r| < 1$∣r∣<1 stated.

Total: 8 marks (M3 A5). Examiners reward candidates who quote the general formulae before substitution and who explicitly verify the convergence condition; both protect a string of accuracy marks even if late arithmetic slips creep in.

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): A geometric series has first term $a$a and common ratio $r$r, with $|r| < 1$∣r∣<1. The sum to infinity is $32$32, and the sum of the squares of the terms is $\dfrac{1024}{3}$31024​.

(a) Show that $\dfrac{a}{1 + r} = \dfrac{32}{3}$1+ra​=332​. (3)

(b) Hence find $a$a and $r$r. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — stating $S_\infty = \dfrac{a}{1 - r} = 32$S∞​=1−ra​=32.
• M1 (AO1.1b) — recognising that the squares form a GP with first term $a^2$a2 and ratio $r^2$r2, hence sum $\dfrac{a^2}{1 - r^2} = \dfrac{1024}{3}$1−r2a2​=31024​.
• A1 (AO2.1) — dividing the second equation by $32$32 (the first), or equivalently using $1 - r^2 = (1 - r)(1 + r)$1−r2=(1−r)(1+r), yields $\dfrac{a}{1 + r} = \dfrac{1024/3}{32} = \dfrac{32}{3}$1+ra​=321024/3​=332​.

(b)

• M1 (AO1.1b) — solving the simultaneous pair $a = 32(1 - r)$a=32(1−r) and $a = \tfrac{32}{3}(1 + r)$a=332​(1+r).
• A1 (AO1.1b) — equating: $32(1 - r) = \tfrac{32}{3}(1 + r)$32(1−r)=332​(1+r), hence $3(1 - r) = 1 + r$3(1−r)=1+r, giving $3 - 3r = 1 + r$3−3r=1+r, $2 = 4r$2=4r, $r = \tfrac{1}{2}$r=21​.
• A1 (AO2.5) — $a = 32(1 - \tfrac{1}{2}) = 16$a=32(1−21​)=16, with check $|r| < 1$∣r∣<1 noted.

Total: 6 marks split AO1 = 4, AO2 = 2. A classic AQA Paper 1 structure: AO1 dominates, with one or two AO2 marks reserved for the synoptic insight that the squares of a GP form another GP — the step that distinguishes A from A* candidates.

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Synoptic links

Connects to:

1. Section E — Exponentials and logarithms: any equation of the form $r^n = k$rn=k with unknown $n$n requires $n = \log_r k = \dfrac{\ln k}{\ln r}$n=logr​k=lnrlnk​. AQA frequently embeds this inside a GP-sum question — "after how many years does the investment first exceed £20{,}000?" — testing logs and series in one step.

2. Section H — Integration: the definite integral $\int_a^b f(x)\,dx$∫ab​f(x)dx is defined as the limit of Riemann sums $\sum_{i=1}^{n} f(x_i^*) \Delta x$∑i=1n​f(xi∗​)Δx. The arithmetic-progression structure of the partition ($x_i = a + i \Delta x$xi​=a+iΔx) and the convergence as $n \to \infty$n→∞ are direct sequences-and-series ideas. Trapezium-rule error analysis sits in this same conceptual space.

3. Section G — Differentiation and Taylor series: Taylor and Maclaurin series express $f(x)$f(x) as an infinite power series $\sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!} x^n$∑n=0∞​n!f(n)(0)​xn. The convergence of these series is a sequences-and-series question; the coefficients come from differentiation. The binomial expansion is the special case for $f(x) = (1 + x)^\alpha$f(x)=(1+x)α.

4. Section J — Numerical methods: fixed-point iteration $x_{n+1} = g(x_n)$xn+1​=g(xn​) is a recurrence relation, and its convergence to a root depends on $|g'(x)| < 1$∣g′(x)∣<1 — a contraction condition that mirrors $|r| < 1$∣r∣<1 for geometric convergence. The same intuition unifies both topics.

5. Section A — Proof: proof by induction on $n$n is the natural tool for verifying closed forms of sums ($\sum_{k=1}^{n} k = \tfrac{n(n+1)}{2}$∑k=1n​k=2n(n+1)​) and for recurrence-defined sequences. AQA Paper 2 routinely sets one induction proof per series, with full marks reserved for clean inductive-step language.

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Mark-scheme literacy

Sequences and series questions on 7357 distribute AO marks more evenly across the three categories than algebra:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Quoting $S_n$Sn​ formulae correctly, computing $n$nth terms from formulae or recurrences, expanding $(1 + x)^n$(1+x)n for small $n$n, applying sigma notation
AO2 (reasoning / interpretation)	25–35%	Justifying convergence ($
AO3 (problem-solving)	10–25%	Modelling savings, depreciation, or population step-change; translating a real-context word problem into AP/GP/recurrence; combined AP+GP simultaneous setups