College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.2 - Arithmetic Sequences - 8.2 Exercises - Page 607: 49


$601$ is the 33rd term of the sequence.

Work Step by Step

RECALL: The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula: $a_n = a + d(n-1)$ where $a$ = first term $d$ = common difference The sequence has $a=25$ and $d=18$. Substituting these tot he formula above gives: $a_n=25 + 18(n-1)$ The $n^{th}$ term is $601$. To know the value of $n$, substitute $601$ to $a_n$ to obtain: $a_n = 25+18(n-1) \\601=25+18(n-1) \\601 - 25 = 18(n-1) \\576 = 18(n-1) \\\dfrac{576}{18}=\dfrac{18(n-1)}{18} \\32=n-1 \\32+1 = n-1+1 \\33=n$ Thus, $601$ is the 33rd term of the sequence.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.