## Algebraic Definition

An $\textit{Arithmetic Sequence}$ is a sequence in which each term differs from the previous one by the same fixed number, 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 can 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} \)

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