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.

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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

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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.

u₂ = 3(4) − 2 = 10
u₃ = 3(10) − 2 = 28
u₄ = 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:

⁷C₃ × 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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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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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

Examiner-rewarded phrasing: "since $|r| = \tfrac{1}{2} < 1$∣r∣=21​<1, the sum to infinity exists and equals $\dfrac{a}{1 - r}$1−ra​"; "the $n$nth term of the AP is $u_n = a + (n - 1)d$un​=a+(n−1)d"; "by induction, assume $P(k)$P(k) holds; show $P(k + 1)$P(k+1) follows". Phrases that lose marks: writing $S_\infty = \dfrac{a}{1 - r}$S∞​=1−ra​ without checking $|r| < 1$∣r∣<1; using $n$n where $n - 1$n−1 is correct in the AP $n$nth term; expanding a binomial $(1 + x)^{1/2}$(1+x)1/2 without stating the validity range $|x| < 1$∣x∣<1.