Answer
The given sequence is neither a geometric sequence nor arithmetic sequence.
Work Step by Step
The given general term is $a_{n}$ = $n^{2}$ - 3
First term $a_{1}$ = $1^{2}$ - 3 = -2 (if n=1)
Second term $a_{2}$ = $2^{2}$ - 3 = 1 (if n=2)
Third term $a_{3}$ = $3^{2}$ - 3 = 6 (if n=3)
Forth term $a_{4}$ = $4^{2}$ - 3 = 13 (if n=4)
Ratio of consecutive terms.
$\frac{13}{6}$ $\ne$ $\frac{6}{1}$$\ne$$\frac{1}{-2}$.
Difference of consecutive terms
(13 - 6 = 7) $\ne$ (6 - 1 = 5) $\ne$ {1 - (-2) = 3}
From above we observe that neither the ratio nor the difference of two consecutive terms is constant. So the given sequence is neither a geometric sequence nor arithmetic sequence.
