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