## Proof That the Set of Arithmetic Sequences is a Vector Space

An arithmetic progression is such that each term is derived from the previous term is by addition of a constant
$d$
.
Suppose we have two arithmetic sequences
$A_1 :a_1, a_1+d_1 , a_1+2d_1 ,..., a_1+nd_1,...$

$A_2 :a_2, a_2+d_2 , a_2+2d_2 ,..., a_2+nd_2,...$

The zero sequence is an arithmetic sequence with first term
$a=0$
and common difference
$d=0$
.
\begin{aligned} & \alpha (a_1, a_1+d_1 , a_1+2d_1 ,..., a_1+nd_1,...) \\ &+ \beta (a_2, a_2+d_2 , a_2+2d_2 ,..., a_2+nd_2,...) \\ &= (\alpha a_1 + \beta a_2 ), (\alpha a_1 + \beta a_2 )+ (\alpha d_1 + \beta d_2 ),(\alpha a_1 + \beta a_2 ) \\ &+ 2(\alpha d_1 + \beta d_2 ) ,...,(\alpha a_1 + \beta a_2 )+ n(\alpha d_1 + \beta d_2 ),... \end{aligned}

which is an arithmetic sequence with first term
$\alpha a_1 + \beta a_2$
and common difference
$\alpha d_1 + \beta d_2$
.
Hence the set of arithmetic sequences forms a vector space.