College Algebra 7th Edition

$n=25$
We are told that we have an arithmetic sequence with: $a=12$ and $d=8$ We know that the sum of such a sequence is: $S_n=\frac{n}{2}[2*a+(n-1)d]$ We with the sum to equal $2700$: $2700=\frac{n}{2}[2*12+(n-1)*8]$ $5400=n(24+8n-8)$ $5400=n(16+8n)$ $5400=16n+8n^2$ $8n^2+16n-5400=0$ $4n^{2}+8n-2700=0$ $4(n+27)(n-25)=0$ $(n+27)=0$ or $(n-25)=0$ $n=-27$ or $n=25$ But $n$ can not be negative, so the answer is $n=25$.