Answer
19 terms
Work Step by Step
We know that $a_n=a_1+(n-1) d$
Here, we have $a_{1}=3; a_n=93; d=5$
Thus, we have
$93=3+(n-1)(5)$
or, $5n-2=93 \implies n=19$
Hence, there are $19$ terms
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 10 - Section 10.2 - Arithmetic Sequences - Exercise Set - Page 1060: 54
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