College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.3 - Page 740: 56

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.
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