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#.
\[1, 2, 3, 4, 5, \ldots, 229, 230\]

#t_1=1#
#d=1#
#n=230#

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 #230# with size #1# of each step, this sequence has #230# terms. Hence, #n=230#.

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?\[a_1=14{\tiny,}\quad a_2=20{\tiny,}\quad a_3=26{\tiny,}\quad a_4=32{\tiny,}\quad\ldots\]

#a_k=# #6 \cdot k +8#

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 #14#.
The difference between the first two terms is #d=20-14=6#.

This means that the direct formula is equal to \(a_k=14+6 \cdot (k-1) \). This can also be written as \[a_k=6 \cdot k + 8\] by expanding the brackets.

New example