## College Algebra (6th Edition)

Total salary over 10 years $S_{10}$ = 5 $\times$ 88500 = 442,500
Starting year's salary $a_{1}$= 33000 Raise in salary per year {common difference (d)} = 2500 Total salary over n years given by sum formula $S_{n}$ = $\frac{n}{2}[ 2 a_{1} + (n - 1)d ]$ Total salary over 10 years $S_{10}$ = $\frac{10}{2}[ 2\times 33000 + (10 - 1)2500 ]$ = $5[ 66000 + 9 \times2500 ]$ = $5[ 66000 + 22500 ]$ = 5 $\times$\times 88500 = 442500