# Arithmetic Sequences

## Algebraic Definition

An $\textit{Arithmetic Sequence}$ is a sequence in which each term differs from the previous one by the same fixed number, which is often called $\textit{common difference}$. It can also be referred to as an $\textit{arithmetic progression}$.

A sequence in mathematics is an ordered set of numbers.
An $\textit{arithmetic sequence}$ is one in which:

• the difference between any two successive terms is the same
• the next term in the sequence is found by adding the same number

For example:

• $1, 3, 5, 7, 9, \cdots$
the common difference is $2$
• $35, 30, 27, 24, \cdots$
the common difference is $-3$

$\{u_{n}\}$ is $\textit{arithmetic}$ if and only if $u_{n+1} – u_{n} = d$ for all positive $n$ where $d$ is a the common difference.

• $3-1=2$
• $5-3=2$
• $7-5=2$
• $u_{n+1} – u_{n} =2$

There the sequence $1, 3, 5, 7, 9, \cdots$ is arithmetic.

## Arithmetic Mean

If $a$, $b$ and $c$ are any consecutive terms of an arithmetic sequence then:

\begin{align} \displaystyle u_{2} – u_{1} &= u_{3} – u_{2} &\text{equating common difference} \\ b-a &= c-b \\ 2b &= a+c \\ \therefore b &= \dfrac{a+c}{2} \\ \end{align}

This means that the middle term is the $\textit{arithmetic mean}$ of the terms on either side of it.

## General Term Formula

Suppose that the first term of an arithmetic sequence is $u_{1}$, or $a$ and the common difference is $d$.

\begin{align} \displaystyle u_{2} &= u_{1} + d \\ &= u_{1} + (2-1)d \\ u_{3} &= u_{2} + d \\ &= (u_{1} + d) + d \\ &= u_{1} + 2d \\ &= u_{1} + (3-1)d \\ u_{4} &= u_{3} + d \\ &= (u_{1} + 2d) +d \\ &= u_{1} + 3d \\ &= u_{1} + (4-1)d \\ &\cdots \\ \therefore u_{n} &= u_{1} + (n-1)d \\ \text{or}\\ \therefore T_{n} &= a + (n-1)d \\ \end{align}

If we are given only two terms of an arithmetic sequence, we are able to use the rule $u_{n}=u_{1}+(n-1)d$ to set up two simultaneous equations to find the value of $u_{1}$, or $a$ and $d$ and hence write down the rule for the arithmetic sequence.

### Example 1

Show that the sequence $3, 10, 17, 24, 31, \cdots$ is arithmetic.

\begin{align} \displaystyle 10 – 3 &= 7 \\ 17 – 10 &= 7 \\ 24 – 17 &= 7 \\ 31 – 24 &= 7 \\ \end{align}
The difference between successive terms is constant.
So the sequence is arithmetic with $u_{1}=3$ and $d=7$.

### Example 2

Find a formula for the general term of $5, 8, 11, 14, 17, \cdots$

\begin{align} \displaystyle 17 – 14 &= 3 \\ 14 – 11 &= 3 \\ 11 – 8 &= 3 \\ 8 – 5 &= 3 \\ \end{align}
The difference between successive terms is constant.
So the sequence is arithmetic with $u_{1}=5$ and $d=3$.
\begin{align} \displaystyle u_{n} &= u_{1} + (n-1) \times d \\ u_{n} &= 5 + (n-1) \times 3 \\ &= 5 + 3n – 3 \\ \therefore u_{n} &= 3n + 2 \\ \end{align}

### Example 3

Find the $150$th term of the sequence: $2, 6, 10, 14, \cdots$.

\begin{align} \displaystyle 14 – 10 &= 4 \\ 10 – 6 &= 4 \\ 6 – 2 &= 4 \\ \end{align}
The difference between successive terms is constant.
So the sequence is arithmetic with $u_{1}=2$ and $d=4$.
\begin{align} \displaystyle u_{n} &= u_{1} + (n-1) \times d \\ u_{n} &= 2 + (n-1) \times 4 \\ &= 2 + 4n – 4 \\ u_{n} &= 4n – 2 \\ u_{150} &= 4 \times 150 – 2 \\ \therefore u_{150} &= 598 \\ \end{align}

### Example 4

Is $83$ a term of the sequence: $4, 7, 10, 13, \cdots$?

\begin{align} \displaystyle 13 – 10 &= 3 \\ 10 – 7 &= 3 \\ 7 – 4 &= 3 \\ \end{align}
The difference between successive terms is constant.
So the sequence is arithmetic with $u_{1}=4$ and $d=3$.
\begin{align} \displaystyle u_{n} &= u_{1} + (n-1) \times d \\ u_{n} &= 4 + (n-1) \times 3 \\ &= 4 + 3n – 3 \\ u_{n} &= 3n + 1 \\ 3n + 1 &= 83 \\ 3n &= 82 \\ n &= 82 \div 3 \\ &= 27.333 \cdots \\ \end{align}
$n$ must be a positive integer, thus $83$ is $\textit{not}$ a term of the sequence.

### Example 5

Which term is $191$ of the sequence $11, 14, 17, 20, \cdots$?

\begin{align} \displaystyle 20 – 17 &= 3 \\ 17 – 14 &= 3 \\ 14 – 11 &= 3 \\ \end{align}
The difference between successive terms is constant.
So the sequence is arithmetic with $u_{1}=11$ and $d=3$.
\begin{align} \displaystyle u_{n} &= u_{1} + (n-1) \times d \\ u_{n} &= 11 + (n-1) \times 3 \\ &= 11 + 3n – 3 \\ u_{n} &= 3n + 8 \\ 3n + 8 &= 191 \\ 3n &= 183 \\ n &= 183 \div 3 \\ &= 61 \\ \end{align}
Therefore $191$ is $61$st term of the sequence.

### Example 6

If $u_{10}=100$ and $u_{15}=175$, find the $n$th term for the arithmetic sequence.

\begin{align} \displaystyle u_{10} &= u_{1} + 9d = 100 \cdots (1) \\ u_{15} &= u_{1} + 14d = 175 \cdots (2) \\ 14d – 9d &= 175 – 100 &(2)-(1) \\ 5d &= 75 \\ d &= 15 \\ u_{1} + 9 \times 15 &= 100 &\text{substitute } d=25 \text{ into } (1) \\ u_{1} + 135 &= 100 \\ u_{1} &= -35 \\ u_{n} &= -35 + (n-1) \times 15 \\ &= -35 + 15n – 15 \\ \therefore u_{n} &= 15n-50 \\ \end{align}