College Algebra (6th Edition)

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.