Answer
Sequence is arithmetic
$d=6$
$a_{n}=a+\left( n-1\right) d=-4+6\left( n-1\right) $
Work Step by Step
In order a sequence to be arithmetic
$a_{n+1}-a_{n}=a_{n}-a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$
$2a_{n}=2\left( 6n-10\right) =12n-20$
$a_{n+1}+a_{n-1}=6\left( n+1\right) -10+6\left( n-1\right) -10=6n+6-10+6n-6-10=12n-20$
$\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$
So this sequence is arithmetic
$a_{n}=a+\left( n-1\right) d=a-d+nd=6n-10\Rightarrow d=6;$
$a-d=-10\Rightarrow a=-4$
So the sequence will be
$a_{n}=a+\left( n-1\right) d=-4+6\left( n-1\right) $