Arithmetic sequences

Sequences and series: Arithmetic sequences and series

Arithmetic sequences

There are two important types of sequences: the arithmetic sequence and the geometric sequence. We will now take a look at the arithmetic sequence.

Arithmetic sequence

An arithmetic sequence is a sequence in which each term is calculated by adding a fixed number to the preceding term.

This fixed number is called the difference of the arithmetic sequence and is often notated with $d$ .

Hence, if $n\gt1$ , then for each $n$ -th term $t_n$ of the sequence we have: $t_n=t_{n-1}+d$

Examples

Give the initial term $t_1$ , the difference $d$ and the number of terms $n$ of the following sequence $t$ .
$1, 2, 3, 4, 5, \ldots, 514, 515$

$t_1=1$
$d=1$
$n=515$

The initial term is the first term and that one is equal to $1$ . Hence, $t_1=1$ .
To calculate the difference, we subtract two subsequent terms from each other: $2-1=1$ . We see that this same difference is also true for other subsequent terms. Hence, it's an arithmetic sequence. In this case, we have $d=1$ .
Since the sequence starts at $1$ and ends at $515$ with size $1$ of each step, this sequence has $515$ terms. Hence, $n=515$ .

New example

The formula in the definition for finding the $n$ -th term is a recursive formula in the sense that the term $t_n$ is given by an expression used in the previous term $t_{n-1}$ . With an arithmetic sequence, we can also construct a direct formula right away. That is a formula for calculating the $n$ -th term with the rank number $n$ , the initial term $t_1$ and the difference $d$ . Here, we do not need the preceding term.

Direct formula arithmetic sequence

The $n$ -th term of an arithmetic sequence $t_1, t_2, \ldots$ satisfies $t_n=t_1+(n-1) \cdot d$ where $d$ is the difference of two subsequent terms in this sequence.

In general, the terms of an arithmetic sequence can be written as follows:

$\begin{array}{lcl} t_1&=&t_1 \\ t_2=t_1+d&=&t_1+d\\ t_3=t_2+d&=&t_1+2d\\ t_4=t_3+d=t_1+2d+d&=&t_1+3d\\ .\\ .\\ .\\ t_{20}=t_{19}+d&=&t_1+19d \end{array}$

This leads to the direct formula: $t_n=t_1+(n-1) \cdot d$

In this case, the sequence starts at $t_1$ . You can also choose to let the sequence start at $t_0$ . In that case, the direct formula is equal to $t_n=t_0+n \cdot d$

This is a result of the fact that $t_n$ now is the $(n+1)$ -th term.

Below are some examples of how we can compose the direct formula and how we can use this one to calculate a term.

Which formula corresponds with the following arithmetic sequence? $a_1=10{\tiny,}\quad a_2=16{\tiny,}\quad a_3=22{\tiny,}\quad a_4=28{\tiny,}\quad\ldots$

$a_k=$ $6 \cdot k +4$

We know that $a_k$ is an arithmetic sequence. According to the direct formula, we have $a_k=a_1+(k-1) \cdot d$ , where $d$ is the difference between two subsequent terms. We have to look for $a_1$ and difference $d$ .

The initial term $a_1$ is given and equal to $10$ .
The difference between the first two terms is $d=16-10=6$ .

This means that the direct formula is equal to $a_k=10+6 \cdot (k-1)$ . This can also be written as $a_k=6 \cdot k + 4$ by expanding the brackets.

New example