Answer
Sequence is arithmetic
$d=\frac{1}{2}$
$\ a_{n}=\dfrac {3}{2}+\dfrac {1}{2}\left( n-1\right) $
Work Step by Step
In order a sequence to be arithmetic
$a_{n+1}-a_{n}=a_{n}-a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$
$2a_{n}=2\left( 1+\dfrac {n}{2}\right) =2+n$
$a_{n+1}+a_{n-1}=1+\dfrac {n+1}{2}+1+\dfrac {n-1}{2}=2+n$
$\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$
So this sequence is arithmetic
$a_{n}=1+\dfrac {n}{2}=a+\left( n-1\right) d=a-d+nd\Rightarrow d=\dfrac {1}{2};a-d=1\Rightarrow a=\dfrac {3}{2}$
$\Rightarrow a_{n}=a+\left( n-1\right) d=\dfrac {3}{2}+\dfrac {1}{2}\left( n-1\right) $
