Answer
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(3(n+1)+1)-(3n+1)=(3n+4)-(3n+1)=3$, thus it is an arithmetic sequence.
$a_1=4$
$a_2=7$
$a_3=10$
$a_4=13$
Work Step by Step
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(3(n+1)+1)-(3n+1)=(3n+4)-(3n+1)=3$, thus it is an arithmetic sequence.
$a_1=3(1)+1=4$
$a_2=3(2)+1=7$
$a_3=3(3)+1=10$
$a_4=3(4)+1=13$
