A-Level Mathematics – Chapter 3: Sequences and Series
Sequences and series appear throughout mathematics and its applications — from financial modelling and population growth to the approximation of complicated functions in physics and engineering. In this chapter you study the two most important families of sequences (arithmetic and geometric) and then the binomial expansion, which provides a systematic way to expand bracketed expressions. The Year 2 binomial section extends this to fractional and negative powers, producing convergent infinite series with important approximation applications.
Specification Note
Content labelled Year 2 is A2-level only and is not required for AS Mathematics. All other content is required for both AS and the full A-Level.
3.1 Arithmetic Sequences and Series
The $n$th Term
Definition — Arithmetic Sequence
An arithmetic sequence (also called an arithmetic progression, AP) is a sequence in which consecutive terms differ by a fixed amount $d$, the common difference. If the first term is $a$, the $n$th term is
$$u_n = a + (n-1)d$$
The sequence is increasing when $d > 0$, decreasing when $d < 0$, and constant when $d = 0$.
Example 3.1 — Finding the $n$th term
An arithmetic sequence has first term $7$ and common difference $-3$. Write down the first five terms and find the $20$th term.
Step 1 First five terms: $7,\ 4,\ 1,\ -2,\ -5$.
Step 2 $u_{20} = 7 + (20-1)(-3) = 7 - 57 = \mathbf{-50}$.
Example 3.2 — Finding $a$ and $d$ from given terms
In an arithmetic sequence, $u_4 = 11$ and $u_9 = 31$. Find $a$ and $d$, then find the value of $n$ for which $u_n = 71$.
Step 1 Set up two equations: $a + 3d = 11$ and $a + 8d = 31$.
Step 2 Subtract: $5d = 20 \implies d = 4$. Then $a = 11 - 12 = -1$.
Step 3 $u_n = 71$: $-1 + (n-1)(4) = 71 \implies 4n = 76 \implies n = 19$.
Sum of an Arithmetic Series
Sum of an Arithmetic Series
The sum of the first $n$ terms of an arithmetic series with first term $a$ and common difference $d$ is
$$S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$$
Equivalently, if the last (i.e.\ $n$th) term is $l$, then $S_n = \dfrac{n}{2}(a + l)$. This second form is convenient when the last term is known directly.
Arithmetic Mean
The arithmetic mean of two numbers $p$ and $q$ is $\dfrac{p+q}{2}$. If $p,\, m,\, q$ are three consecutive terms of an AP, then $m = \dfrac{p+q}{2}$.
Example 3.3 — Sum of a given arithmetic series
Find the sum of the series $5 + 8 + 11 + \cdots + 50$.
Step 1 $a = 5$, $d = 3$, $l = 50$. Number of terms: $50 = 5 + (n-1)(3) \implies n = 16$.
Step 2 $S_{16} = \dfrac{16}{2}(5 + 50) = 8 \times 55 = \mathbf{440}$.
Example 3.4 — Deriving $a$ and $d$ from $S_n$
The sum of the first $n$ terms of an arithmetic series is $S_n = n^2 + 4n$. Find the first term and the common difference.
Step 1 $a = u_1 = S_1 = 1 + 4 = 5$.
Step 2 $u_2 = S_2 - S_1 = (4+8) - 5 = 7$. So $d = 7 - 5 = \mathbf{2}$.
Check: $S_n = \dfrac{n}{2}[10 + (n-1)(2)] = n^2 + 4n$ ✓
Example 3.5 — Setting up and solving simultaneous equations
The 4th term of an AP is $13$ and the sum of the first 10 terms is $175$. Find $a$ and $d$.
Step 1 $a + 3d = 13$ (i)
Step 2 $S_{10} = \dfrac{10}{2}(2a + 9d) = 175 \implies 2a + 9d = 35$ (ii)
Step 3 From (i): $a = 13 - 3d$. Substitute into (ii): $2(13-3d) + 9d = 35 \implies 3d = 9 \implies d = 3,\; a = 4$.
Exam Tip — Arithmetic Series
Use $S_n = \frac{n}{2}(a+l)$ when you know the first and last terms, and $S_n = \frac{n}{2}[2a+(n-1)d]$ otherwise. Always identify $a$, $d$, and $n$ before substituting. When given $S_n$ as a formula, compute $u_n = S_n - S_{n-1}$ to find the $n$th term.
3.2 Geometric Sequences and Series
The $n$th Term
Definition — Geometric Sequence
A geometric sequence (also called a geometric progression, GP) is a sequence in which each term is obtained by multiplying the previous one by a fixed number $r$, the common ratio. If the first term is $a$, the $n$th term is
$$u_n = a r^{n-1}$$
The ratio can be recovered as $r = \dfrac{u_{n+1}}{u_n}$.
Figure 3.1 — Terms of a geometric sequence plotted as points. Use the sliders to adjust $a$ (first term) and $r$ (common ratio). Observe that $|r| > 1$ produces unbounded growth or oscillation, while $|r| < 1$ produces decay towards zero.
Example 3.6 — $n$th term and first term exceeding a bound
A geometric sequence has first term $3$ and common ratio $2$. Find the 8th term and the first term greater than $500$.
Step 1 $u_8 = 3 \times 2^7 = 3 \times 128 = \mathbf{384}$.
Step 2 $3 \times 2^{n-1} > 500 \implies 2^{n-1} > 166.\overline{6}$. Since $2^7 = 128 < 166.\overline{6}$ and $2^8 = 256 > 166.\overline{6}$, we need $n - 1 = 8$, i.e.\ $n = 9$. Check: $u_9 = 3 \times 256 = 768$ ✓
Sum of a Geometric Series
Sum of a Finite Geometric Series
The sum of the first $n$ terms of a geometric series is
$$S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \ne 1)$$
An equivalent form, convenient when $|r| > 1$, is $S_n = \dfrac{a(r^n - 1)}{r - 1}$.
Sum to Infinity of a Geometric Series
When $|r| < 1$, the partial sums converge to a finite limit:
$$S_\infty = \frac{a}{1-r}, \quad |r| < 1$$
If $|r| \ge 1$, the series diverges and no finite sum to infinity exists.
Geometric Mean
The geometric mean of two positive numbers $p$ and $q$ is $\sqrt{pq}$. If $p,\, m,\, q$ are three consecutive terms of a GP, then $m^2 = pq$.
Example 3.7 — Sum of a finite geometric series
Find the sum of the first $10$ terms of the series $2 + 6 + 18 + \cdots$
Step 1 $a = 2$, $r = 3$.
Step 2 $S_{10} = \dfrac{2(3^{10}-1)}{3-1} = \dfrac{2(59048)}{2} = \mathbf{59048}$.
Example 3.8 — Sum to infinity and convergence
A geometric series has first term $12$ and common ratio $\tfrac{1}{4}$. Find the sum to infinity and the least $n$ for which $S_n$ exceeds $15.9$.
Step 1 $S_\infty = \dfrac{12}{1 - \frac{1}{4}} = 16$.
Step 2 $S_n = 16\!\left(1 - \left(\tfrac{1}{4}\right)^n\right) > 15.9 \implies \left(\tfrac{1}{4}\right)^n < 0.00625$.
Take logs: $n\ln(\tfrac{1}{4}) < \ln(0.00625) \implies n > \dfrac{\ln(0.00625)}{\ln(0.25)} \approx 3.66$. Least integer: $n = \mathbf{4}$.
Example 3.9 — Compound interest application
A savings account starts with £500 and earns $3\%$ compound interest per year. Write down the value after $n$ years as a geometric sequence, and find the first year in which the value exceeds £700.
Step 1 $V_n = 500 \times 1.03^n$ (GP with $a = 500$, $r = 1.03$).
Step 2 $500 \times 1.03^n > 700 \implies 1.03^n > 1.4 \implies n > \dfrac{\ln 1.4}{\ln 1.03} \approx 11.38$. First year: $n = \mathbf{12}$.
Exam Tip — Geometric Series
Always check that $|r| < 1$ before applying the sum-to-infinity formula. Applying it when $|r| \ge 1$ gives a nonsensical result and will score zero for that part. Note also that $r$ may be negative: $r = -\tfrac{1}{2}$ still satisfies $|r| < 1$, so the sum to infinity exists.
3.3 Binomial Expansion (Integer Powers)
Pascal’s Triangle
Pascal’s triangle arranges the binomial coefficients for each power $n$. Each entry is the sum of the two entries immediately above it. The rows correspond to $n = 0, 1, 2, \ldots$
Pascal’s Triangle ($n = 0$ to $5$)
The Binomial Theorem
Binomial Theorem (Positive Integer Powers)
For any positive integer $n$:
$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$
where the binomial coefficient is $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$, also written ${}^nC_r$.
The general term (the $(r+1)$th term) is $\dbinom{n}{r}a^{n-r}b^r$.
Key symmetry: $\dbinom{n}{r} = \dbinom{n}{n-r}$.
Example 3.10 — Full expansion of $(a+b)^4$
Expand $(2x + 3)^4$ fully.
Step 1 Apply the theorem with $a = 2x$, $b = 3$, $n = 4$, using the $n = 4$ row of Pascal’s triangle (coefficients $1, 4, 6, 4, 1$):
$$1\cdot(2x)^4 + 4\cdot(2x)^3(3) + 6\cdot(2x)^2(3)^2 + 4\cdot(2x)(3)^3 + 1\cdot(3)^4$$
Step 2 Evaluate:
$$= 16x^4 + 4 \cdot 8x^3 \cdot 3 + 6 \cdot 4x^2 \cdot 9 + 4 \cdot 2x \cdot 27 + 81$$
$$= \mathbf{16x^4 + 96x^3 + 216x^2 + 216x + 81}$$
Example 3.11 — Finding a specific term
Find the coefficient of $x^3$ in the expansion of $(1 - 2x)^7$.
Step 1 General term: $\dbinom{7}{r}(1)^{7-r}(-2x)^r = \dbinom{7}{r}(-2)^r x^r$.
Step 2 For $x^3$, set $r = 3$: $\dbinom{7}{3}(-2)^3 = 35 \times (-8) = \mathbf{-280}$.
Example 3.12 — Two-bracket expansion
Find the coefficient of $x^2$ in the expansion of $(1 + x)^5(1 - 3x)$.
Step 1 Expand $(1+x)^5$ up to $x^2$: $(1+x)^5 = 1 + 5x + 10x^2 + \cdots$
Step 2 Multiply by $(1-3x)$ and collect terms in $x^2$:
$$1 \cdot 10x^2 + (-3) \cdot 5x^2 = 10x^2 - 15x^2 = \mathbf{-5x^2}$$
Coefficient is $-5$.
Example 3.13 — Binomial approximation
Use the expansion of $(1+x)^5$ to find an approximation for $1.02^5$.
Step 1 Write $1.02^5 = (1 + 0.02)^5$ and expand for small $x = 0.02$:
$$(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + \cdots$$
Step 2 Substitute: $\approx 1 + 0.1 + 10(0.0004) + 10(0.000008) = 1 + 0.1 + 0.004 + 0.00008 = \mathbf{1.10408}$.
(Exact: $1.104080\overline{32}$ — the approximation is accurate to 6 significant figures.)
Exam Tip — Binomial Expansion (Integer Powers)
Always keep track of which power of $b$ corresponds to the index $r$ in $\binom{n}{r}$. When $b$ is negative (e.g.\ $-2x$), expand $(-2x)^r$ separately before multiplying: $(-2x)^3 = -8x^3$, not $+8x^3$. Check your expansion by verifying that the coefficients sum to $2^n$.
3.4 Binomial Expansion (General) Year 2
The General Binomial Series
For positive integer $n$, the binomial expansion terminates after $n+1$ terms. For any rational $n$ (including fractions and negative values), the expansion becomes an infinite series that converges only when $|x| < 1$.
General Binomial Series
For any rational $n$ and $|x| < 1$:
$$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$
The series is infinite when $n$ is not a positive integer. The coefficient of $x^r$ in the $(r+1)$th term is $\dfrac{n(n-1)\cdots(n-r+1)}{r!}$.
Validity Condition
The general binomial series $(1+x)^n$ is valid for $\mathbf{|x| < 1}$. For expressions of the form $(a + bx)^n$, factorise first:
$$(a + bx)^n = a^n\!\left(1 + \frac{bx}{a}\right)^{\!n}$$
and then apply the series with the condition $\left|\dfrac{bx}{a}\right| < 1$, i.e.\ $|x| < \dfrac{|a|}{|b|}$.
Figure 3.2 — The function $y = (1+x)^{1/2}$ (red) compared with the 1-term, 2-term, and 3-term binomial approximations (blue shades). The approximations are accurate near $x = 0$ but diverge outside $|x| < 1$ (dashed vertical lines).
Example 3.14 — Expansion for a fractional power
Expand $(1+x)^{1/2}$ up to and including the $x^3$ term. State the range of validity.
Step 1 Apply the general formula with $n = \tfrac{1}{2}$:
$$\text{Term 1: } 1$$
$$\text{Term 2: } \tfrac{1}{2}x$$
$$\text{Term 3: } \frac{\frac{1}{2}\!\left(-\frac{1}{2}\right)}{2!}x^2 = -\frac{1}{8}x^2$$
$$\text{Term 4: } \frac{\frac{1}{2}\!\left(-\frac{1}{2}\right)\!\left(-\frac{3}{2}\right)}{3!}x^3 = \frac{3}{48}x^3 = \frac{1}{16}x^3$$
Step 2 $(1+x)^{1/2} \approx 1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \dfrac{x^3}{16}$, valid for $|x| < 1$.
Example 3.15 — Expansion for a negative power
Find the first three terms of the expansion of $\dfrac{1}{(1+2x)^3}$ and state the range of validity.
Step 1 Write as $(1+2x)^{-3}$ with $n = -3$ and $x \mapsto 2x$:
$$1 + (-3)(2x) + \frac{(-3)(-4)}{2!}(2x)^2 + \cdots = 1 - 6x + 6 \cdot 4x^2 + \cdots$$
Step 2 $(1+2x)^{-3} \approx \mathbf{1 - 6x + 24x^2} + \cdots$
Step 3 Validity: $|2x| < 1 \implies |x| < \dfrac{1}{2}$.
Example 3.16 — Handling $(a + bx)^n$
Find the first three terms of $(4-x)^{1/2}$ in ascending powers of $x$, and state the range of validity.
Step 1 Factorise: $(4-x)^{1/2} = 2\!\left(1 - \dfrac{x}{4}\right)^{\!1/2}$.
Step 2 Expand $\left(1 - \dfrac{x}{4}\right)^{\!1/2}$ using Example 3.14 with $x \mapsto -\dfrac{x}{4}$:
$$\approx 1 + \frac{1}{2}\!\left(-\frac{x}{4}\right) - \frac{1}{8}\!\left(\frac{x}{4}\right)^{\!2} = 1 - \frac{x}{8} - \frac{x^2}{128}$$
Step 3 Multiply by $2$: $(4-x)^{1/2} \approx \mathbf{2 - \dfrac{x}{4} - \dfrac{x^2}{64}}$.
Validity: $\left|\dfrac{x}{4}\right| < 1 \implies |x| < 4$.
Example 3.17 — Approximation using the binomial series
Use the expansion of $(1+x)^{-1}$ with a suitable value of $x$ to find an approximation for $\dfrac{1}{1.05}$.
Step 1 $(1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots$, valid for $|x| < 1$.
Step 2 Substitute $x = 0.05$:
$$\frac{1}{1.05} \approx 1 - 0.05 + 0.0025 - 0.000125 = 0.952375$$
Exact value: $\dfrac{1}{1.05} = 0.9523\overline{809523}$. The approximation is accurate to 5 significant figures.
Exam Tip — General Binomial Expansion
Never omit the validity condition $|x| < 1$ (or its modified form) — this is routinely penalised. Equally important: when dealing with $(a + bx)^n$, always factorise to extract $a^n$ before expanding. Omitting the $a^n$ factor is the single most common error in this topic.
Practice Problems
Problem 1
An arithmetic sequence has first term $5$ and $25$th term $101$. Find the common difference and the sum of the first $25$ terms.
Show solution
$u_{25} = 5 + 24d = 101 \implies d = 4$.
$S_{25} = \dfrac{25}{2}(5 + 101) = \dfrac{25 \times 106}{2} = \mathbf{1325}$.
Problem 2
The sum of the first $n$ terms of an arithmetic series is $S_n = \dfrac{n}{2}(3n+11)$. Find the 10th term and show that all terms are positive.
Show solution
$a = S_1 = 7$. $u_2 = S_2 - S_1 = 17 - 7 = 10$. So $d = 3$.
$u_{10} = 7 + 9 \times 3 = \mathbf{34}$.
$u_n = 7 + (n-1)(3) = 3n + 4 > 0$ for all positive integers $n$ ✓
Problem 3
A geometric sequence has second term $6$ and fifth term $\dfrac{3}{32}$. Find the common ratio and the sum to infinity.
Show solution
$u_2 = ar = 6$ and $u_5 = ar^4 = \dfrac{3}{32}$. Divide: $r^3 = \dfrac{3/32}{6} = \dfrac{1}{64} \implies r = \dfrac{1}{4}$.
$a = \dfrac{6}{1/4} = 24$. Since $|r| < 1$: $S_\infty = \dfrac{24}{1-\frac{1}{4}} = \mathbf{32}$.
Problem 4
A geometric series has first term $a > 0$ and common ratio $r$ where $0 < r < 1$. The sum of the first three terms is $\dfrac{7}{8}$ times the sum to infinity. Show that $r^3 = \dfrac{1}{8}$ and hence find $r$.
Show solution
$S_3 = \dfrac{a(1-r^3)}{1-r}$; $S_\infty = \dfrac{a}{1-r}$.
$S_3 = \dfrac{7}{8}S_\infty \implies \dfrac{a(1-r^3)}{1-r} = \dfrac{7}{8} \cdot \dfrac{a}{1-r}$.
Cancel $\dfrac{a}{1-r}$: $1 - r^3 = \dfrac{7}{8} \implies r^3 = \dfrac{1}{8} \implies r = \dfrac{1}{2}$ ✓
Problem 5
Find the coefficient of $x^4$ in the expansion of $(3 - 2x)^6$.
Show solution
General term: $\dbinom{6}{r}(3)^{6-r}(-2x)^r = \dbinom{6}{r}3^{6-r}(-2)^r x^r$.
For $x^4$, $r = 4$: $\dbinom{6}{4} \cdot 3^2 \cdot (-2)^4 = 15 \times 9 \times 16 = \mathbf{2160}$.
Problem 6
Given that the expansion of $(1 + kx)^6$ starts $1 + 18x + ax^2 + \cdots$, find $k$ and $a$.
Show solution
Coefficient of $x$: $\dbinom{6}{1}k = 6k = 18 \implies k = 3$.
Coefficient of $x^2$: $\dbinom{6}{2}k^2 = 15 \times 9 = \mathbf{135}$. So $a = 135$.
Problem 7
Find the coefficient of $x^2$ in the expansion of $(2 + x)^5(1 - 3x)$.
Show solution
$(2+x)^5 = 32 + 80x + 80x^2 + \cdots$
Coefficient of $x^2$ in $(2+x)^5(1-3x)$: $80 \cdot 1 + 80 \cdot (-3) = 80 - 240 = \mathbf{-160}$.
Problem 8 Year 2
Find the first three terms of $\dfrac{1}{\sqrt{1-4x}}$ in ascending powers of $x$ and state the range of validity.
Show solution
$(1-4x)^{-1/2}$ with $n = -\tfrac{1}{2}$, $x \mapsto -4x$:
$$1 + \left(-\tfrac{1}{2}\right)(-4x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(-4x)^2 + \cdots = 1 + 2x + 6x^2 + \cdots$$
Validity: $|{-4x}| < 1 \implies |x| < \dfrac{1}{4}$.
Problem 9 Year 2
Find the first three terms of the expansion of $(9 + x)^{1/2}$ in ascending powers of $x$, and use an appropriate substitution to find an approximation for $\sqrt{10}$, giving your answer to 4 decimal places.
Show solution
$(9+x)^{1/2} = 3\!\left(1 + \dfrac{x}{9}\right)^{\!1/2} \approx 3\!\left(1 + \dfrac{x}{18} - \dfrac{x^2}{648}\right) = 3 + \dfrac{x}{6} - \dfrac{x^2}{216}$. Valid for $|x| < 9$.
Substitute $x = 1$: $\sqrt{10} \approx 3 + \dfrac{1}{6} - \dfrac{1}{216} = 3 + 0.16\overline{6} - 0.004629\ldots \approx \mathbf{3.1620}$.
(Calculator: $\sqrt{10} = 3.16227\ldots$ — correct to 4 d.p.)
Problem 10 Year 2
The expansion of $(1 + ax)^n$ in ascending powers of $x$ begins $1 - 12x + 60x^2 - \cdots$ Find $a$ and $n$, and state the range of validity.
Show solution
Coefficient of $x$: $na = -12$. Coefficient of $x^2$: $\dfrac{n(n-1)}{2}a^2 = 60$.
From the first: $a = -12/n$. Substitute into the second:
$\dfrac{n(n-1)}{2} \cdot \dfrac{144}{n^2} = 60 \implies \dfrac{144(n-1)}{2n} = 60 \implies 144n - 144 = 120n \implies 24n = 144 \implies n = 6$.
Then $a = -2$. Since $n = 6$ is a positive integer, the series terminates and is valid for all $x$. (Equivalently, $(1-2x)^6$, which is a finite polynomial.)