## 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 \[d=0\]

.\[\begin{equation} \begin{aligned} & \alpha (a_1, a_1+d_1 , a_1+2d_1 ,..., a_1+nd_1,...) \\ &+ \beta (a_2, a_2+d_2 , a_2+2d_2 ,..., a_2+nd_2,...) \\ &= (\alpha a_1 + \beta a_2 ), (\alpha a_1 + \beta a_2 )+ (\alpha d_1 + \beta d_2 ),(\alpha a_1 + \beta a_2 ) \\ &+ 2(\alpha d_1 + \beta d_2 ) ,...,(\alpha a_1 + \beta a_2 )+ n(\alpha d_1 + \beta d_2 ),... \end{aligned} \end{equation}\]

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.