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: 65



Work Step by Step

We are told that we have an arithmetic sequence with: $a=5$ and $d=2$ We know that the sum of an arithmetic sequence is given by: $S_{n}= \frac{n}{2}[2a+(n-1)d]$ We need this to equal $2700$: $2700= \frac{n}{2}[10+2(n-1)]$ $5400=10n+2n^{2}-2n$ $10n+2n^{2}-2n-5400=0$ $2n^{2}+8n-5400=0$ $n^{2}+4n-2700=0$ $(n-50)(n+54)=0$ $(n-50)=0$ or $(n+54)=0$ $n=50$ or $n=-54$ But $n$ can not be negative, so the answer is $n=50$.
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