Answer
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(6-2(n+1))-(6-2n)=(4-2n)-(6-2n)=-2$, thus it is an arithmetic sequence.
$a_1=4$
$a_2=2$
$a_3=0$
$a_4=-2$
Work Step by Step
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(6-2(n+1))-(6-2n)=(4-2n)-(6-2n)=-2$, thus it is an arithmetic sequence.
$a_1=6-2(1)=4$
$a_2=6-2(2)=2$
$a_3=6-2(3)=0$
$a_4=6-2(4)=-2$
