## 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 $a=12, d=8,$ $a_{n}=a+(n-1)d =12+8(n-1)=8n+4$ Also, given that $S_{n}=2700=n(\displaystyle \frac{a+a_{n}}{2})$ we find n: $2700=n(\displaystyle \frac{12+8n+4}{2})$ $2700=n(\displaystyle \frac{8n+16}{2})$ $2700=n(4n+8)$ $4n^{2}+8n-2700=0\qquad /\div 4$ $n^{2}+2n-675=0$ Quadratic formula: $n=\displaystyle \frac{-2\pm\sqrt{4+2700}}{2}=\frac{-2\pm 54}{2}$ $n=-1\pm 27$ ...discard the negative solution, n=26 There are 26 terms in the sum.