Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1074: 55



Work Step by Step

We are given that $ a_n=n^2+5$ Here, we have the first four terms $ a_1=6, a_2=9,a_3=14,a_4=21$ when $ n=1,2,3,4$ Now, subtract each two consecutive terms as below: $ a_2-a_1=9-6=3; a_3-a_2=14-7=5; a_4-a_3=21-14=7$ and $ a_2/a_1=\dfrac{9}{6}=\dfrac{3}{2}; a_3/a_2=\dfrac{14}{9}; a_4/a_3=\dfrac{21}{14}=\dfrac{3}{2}$ It has been seen that we did not get the same constant, so the sequence is neither a geometric sequence nor an arithmetic sequence.
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