## Precalculus: Mathematics for Calculus, 7th Edition

See p. 855. For the arithmetic sequence $a_{n}=a+(n-1)d$ the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$ is given by either of the following equivalent formulas: 1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad$2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$ ------------------- Given $a=5, d=2700,$ $a_{n}=a+(n-1)d =5+2(n-1).$ Also, given that $S_{n}=2700=\displaystyle \frac{n}{2}[2a+(n-1)d]$, we find n: $2700=\displaystyle \frac{n}{2}[2(5)+2(n-1)]\quad/\times 2$ $5400=n(2n+8)$ $2n^{2}+8n-5400=0\qquad/\div 2$ $n^{2}+4n-2700=0$ ... factoring: two factors of $-2700$ that add to 4 are $-54$ and 50. (or , use the quadratic formula) $(n+50)(n-50)=0$ $n=50$ (discard the negative solution) There are 50 terms in the sum.