🔢 Sequences and Series Overview

A Level Maths Year 2 · Overview of the whole topic

Level 3 · Ages 17–18
1 The Big Idea

A sequence is a list of numbers that follows a rule. A series is what you get when you add the terms up. This topic (DfE content area D · Sequences and series) is about describing sequences precisely, and about finding sums without adding every term by hand.

Two families dominate. In an arithmetic sequence you add the same amount each time; in a geometric sequence you multiply by the same amount each time. Each family has a formula for its $n$th term and a formula for its sum — and you are expected to be able to prove both sum formulae.
What is new compared with Year 1
Recurrence relations and sigma notation · increasing, decreasing and periodic sequences
Arithmetic and geometric series · the sum to infinity
The binomial expansion of $(a + bx)^n$ for any rational $n$
Calculators are allowed, and they make light work of generating terms of a recurrence. But a question that says "show that" or asks for a proof needs every algebraic step written out — a list of numbers checked on a calculator is not a proof.
2 The Language You Need
WordMeaningExample
TermOne number in the sequence; $u_n$ is the $n$th term.In $2, 5, 8$: $u_2 = 5$
Recurrence relationA rule giving each term from the previous one.$u_{n+1} = 2u_n + 1$, $u_1 = 3$
SeriesThe sum of the terms of a sequence.$2 + 5 + 8 + 11$
$S_n$The sum of the first $n$ terms.$S_3 = 2 + 5 + 8 = 15$
$\Sigma$ (sigma)"Add up", with limits for the dummy variable.$\sum_{r=1}^{4} r^2 = 30$
Common difference $d$The fixed amount added in an arithmetic sequence.$d = 3$ in $2, 5, 8$
Common ratio $r$The fixed multiplier in a geometric sequence.$r = \tfrac12$ in $8, 4, 2$
ConvergentThe terms (or sums) approach a fixed limit.$S_n$ for $1 + \tfrac12 + \tfrac14 + \cdots \to 2$
DivergentNot convergent: no finite limit.$1 + 2 + 4 + 8 + \cdots$
PeriodicThe terms repeat in a cycle; the cycle length is the order.$1, 3, 1, 3, \ldots$ has order $2$
Sequence or series? A sequence is a list, separated by commas. A series is a sum, joined by plus signs. "Find the $10$th term" is about the sequence; "find the sum of the first $10$ terms" is about the series.
3 The Core Formulae
ArithmeticGeometric
Rule$u_{n+1} = u_n + d$$u_{n+1} = r\,u_n$
$n$th term$a + (n-1)d$$ar^{n-1}$
Sum of $n$ terms$\tfrac{n}{2}[2a + (n-1)d]$ or $\tfrac{n}{2}(a + l)$$\dfrac{a(1 - r^n)}{1 - r}$,  $r \neq 1$
Sum to infinityNever exists (unless $a = d = 0$)$\dfrac{a}{1-r}$, only if $|r| < 1$
Proof of the sumWrite forwards and backwards, addWrite $S_n$ and $rS_n$, subtract
The general binomial expansion — valid for $|x| < 1$
$(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots$
Worked Example — An arithmetic and a geometric sequence together

An arithmetic sequence has first term $a$ and non-zero common difference $d$. Its $1$st, $3$rd and $9$th terms are the first three terms of a geometric sequence. (a) Show that $d = a$. (b) Given $a = 2$, find the sum of the first $20$ terms of the arithmetic sequence and the sum of the first $5$ terms of the geometric sequence.

①The terms are $a$, $a + 2d$, $a + 8d$. For a geometric sequence, $(\text{middle})^2 = \text{first} \times \text{third}$.
②$(a + 2d)^2 = a(a + 8d) \Rightarrow a^2 + 4ad + 4d^2 = a^2 + 8ad \Rightarrow 4d^2 = 4ad$.
③$d \neq 0$, so divide by $4d$: $d = a$. $\blacksquare$
④With $a = d = 2$: $S_{20} = \tfrac{20}{2}[4 + 19 \times 2] = 10 \times 42 = 420$.
⑤The geometric terms are $2, 6, 18$, so $r = 3$: $S_5 = \dfrac{2(3^5 - 1)}{3 - 1} = 242$.
The statement "$d$ is non-zero" is there for a reason: it is what allows you to divide by $d$. Always say so when you use it.
4 The Sequences and Series Subtopics

All of DfE content area D, and the page that covers each part.

5 Common Mistakes to Avoid
Using $n$ instead of $n - 1$. The $n$th term is $a + (n-1)d$ or $ar^{n-1}$. The first term has no steps taken, so it uses $0$, not $1$.
Miscounting terms. From $r = 5$ to $r = 20$ there are $16$ terms, not $15$. From $102$ to $297$ in steps of $3$ there are $\tfrac{297-102}{3} + 1 = 66$ terms.
Using $S_\infty$ when it does not exist. $\dfrac{a}{1-r}$ is only valid for $|r| < 1$. For $r = 2$ it gives a negative "sum" of positive numbers — a sure sign of an error.
Forgetting to reverse an inequality. Solving $0.9^n < 0.2$ by logs means dividing by $\ln 0.9$, which is negative. The answer is $n > \dfrac{\ln 0.2}{\ln 0.9}$.
Binomial errors. In $(2 + 3x)^{-2}$ you must take out $2^{-2} = \tfrac14$, not $2$. In $(1 - 2x)^{1/2}$ the square of $-2x$ is $4x^2$, not $-2x^2$. And always state the range of validity.
6 Quick Reference

Recurrence

$u_{n+1} = \mathrm{f}(u_n)$; a limit satisfies $L = \mathrm{f}(L)$.

Sigma

$\sum_{r=a}^{b}$ has $b - a + 1$ terms.

Arithmetic

$u_n = a + (n-1)d$,  $S_n = \tfrac{n}{2}(a + l)$

Natural numbers

$\sum_{r=1}^{n} r = \tfrac12n(n+1)$

Geometric

$u_n = ar^{n-1}$,  $S_n = \dfrac{a(1-r^n)}{1-r}$

Sum to infinity

$\dfrac{a}{1-r}$ for $|r| < 1$

Consecutive terms

AP: $2b = a + c$.  GP: $b^2 = ac$.

Binomial

$(1+bx)^n$ valid for $|bx| < 1$.

Finding $n$

Quadratic for an AP; logarithms for a GP.

7 Practice Questions

Ten questions drawing on the whole of Sequences and Series. Work each one through before opening the solution.

Question 1

Write down the first four terms of the sequence with $n$th term $u_n = 2^n - n$.

▶ Show solution

$u_1 = 2 - 1 = 1$, $u_2 = 4 - 2 = 2$, $u_3 = 8 - 3 = 5$, $u_4 = 16 - 4 = 12$.

Question 2

Evaluate $\displaystyle\sum_{r=1}^{12} (2r + 5)$.

▶ Show solution

$2\displaystyle\sum_{r=1}^{12} r + \sum_{r=1}^{12} 5 = 2 \times \tfrac12(12)(13) + 60 = 156 + 60 = 216$.

Question 3

An arithmetic sequence has first term $-7$ and common difference $3$. Find the $20$th term and the sum of the first $20$ terms.

▶ Show solution

$u_{20} = -7 + 19 \times 3 = 50$.

$S_{20} = \tfrac{20}{2}(-7 + 50) = 10 \times 43 = 430$.

Question 4

Find the sum to infinity of $6 + 4 + \tfrac83 + \cdots$

▶ Show solution

$r = \tfrac46 = \tfrac23$, and $|r| < 1$.

$S_\infty = \dfrac{6}{1 - \frac23} = 18$.

Question 5

$u_{n+1} = 4 - u_n$ with $u_1 = 1$. Describe the sequence and find $\displaystyle\sum_{r=1}^{51} u_r$.

▶ Show solution

$u_2 = 3$, $u_3 = 1$, $u_4 = 3, \ldots$ — periodic with order $2$.

$51 = 2 \times 25 + 1$: twenty-five pairs each summing to $4$, plus $u_{51} = 1$.

Sum $= 100 + 1 = 101$.

Question 6

Expand $(1 + x)^{1/3}$ in ascending powers of $x$ up to $x^2$, stating the range of validity.

▶ Show solution

$1 + \tfrac13x + \dfrac{\frac13\left(-\frac23\right)}{2}x^2 = 1 + \tfrac13x - \tfrac19x^2 + \cdots$

Valid for $|x| < 1$.

Question 7

The $3$rd term of an arithmetic series is $9$ and the sum of the first $10$ terms is $165$. Find the first term and the common difference.

▶ Show solution

$a + 2d = 9$ and $\tfrac{10}{2}(2a + 9d) = 165 \Rightarrow 2a + 9d = 33$.

Double the first: $2a + 4d = 18$. Subtract: $5d = 15$, so $d = 3$ and $a = 3$.

Question 8

A geometric series has first term $8$ and fourth term $1$. Find the sum to infinity, and the least $n$ for which $S_n > 15.9$.

▶ Show solution

$8r^3 = 1 \Rightarrow r = \tfrac12$.  $S_\infty = \dfrac{8}{1 - \frac12} = 16$.

$S_n = 16\left(1 - 0.5^n\right) > 15.9 \Rightarrow 0.5^n < \tfrac{0.1}{16} = \tfrac{1}{160}$.

$n > \dfrac{\ln(1/160)}{\ln 0.5} = 7.32\ldots$ (reversed, as $\ln 0.5 < 0$), so $n = 8$.

Question 9

Find $\displaystyle\sum_{r=1}^{\infty} 5(0.4)^r$.

▶ Show solution

The first term ($r = 1$) is $5 \times 0.4 = 2$, and the ratio is $0.4$.

$S_\infty = \dfrac{2}{1 - 0.4} = \dfrac{2}{0.6} = \dfrac{10}{3}$.

Question 10

A graduate is offered two jobs, each with a starting salary of £$30\,000$. In Job A the salary rises by £$1200$ each year. In Job B it rises by $3.5\%$ each year.

(a) Find the salary in year $10$ for each job.   (b) Find the total earned over the first $10$ years in each job.   (c) Find the first year in which the Job B salary is greater than the Job A salary.

▶ Show solution

(a) A (arithmetic): $30\,000 + 9 \times 1200 = $ £$40\,800$.  B (geometric): $30\,000 \times 1.035^9 = $ £$40\,886.92$.

(b) A: $S_{10} = \tfrac{10}{2}(30\,000 + 40\,800) = $ £$354\,000$.

B: $S_{10} = \dfrac{30\,000(1.035^{10} - 1)}{0.035} = $ £$351\,941.79$.

Job A pays more in total over ten years, even though Job B's salary is higher by year $10$.

(c) Need $30\,000 \times 1.035^{n-1} > 30\,000 + 1200(n-1)$. This mixes a power with a linear term, so test values:

$n = 9$: A £$39\,600$, B £$39\,504.27$ (A higher).  $n = 10$: A £$40\,800$, B £$40\,886.92$ (B higher).

Job B's salary first exceeds Job A's in year 10.

Sequences and Series Overview · A Level Maths Year 2 · Created with MathJax