## College Algebra 7th Edition

$n=50$
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 $n=50$.