Answer
Neither
Work Step by Step
We are given that $ a_n=n^2-3$
Here, we have the first four terms $ a_1=-2, a_2=1,a_3=6,a_4=13$ when $ n=1,2,3,4$
Now, subtract each two consecutive terms as below: $ a_2-a_1=1+2=3; a_3-a_2=6-1=5; a_4-a_3=13-6=7$
and $ a_2/a_1=\dfrac{-1}{2}; a_3/a_2=6; a_4/a_3=\dfrac{13}{6}$
It has been seen that we did not get the same constant, so the sequence is neither geometric nor arithmetic.
