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: 50

Answer

$2795$ is the 48th 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 $a=3500$ and $d=-15$. Substituting these tot he formula above gives: $a_n=3500 + (-15)(n-1) \\a_n=3500-15(n-1)$ The $n^{th}$ term is $2795$. To know the value of $n$, substitute $2795$ to $a_n$ to obtain: $a_n = 3500-15(n-1) \\2795=3500-15(n-1) \\2795 - 3500 = -15(n-1) \\-705 = -15(n-1) \\\dfrac{-705}{-15}=\dfrac{-15(n-1)}{-15} \\47=n-1 \\47+1 = n-1+1 \\48=n$ Thus, $2795$ is the 48th term of the sequence.
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