## College Algebra (10th Edition)

$d=-\displaystyle \frac{1}{3},$ which is constant, so the sequence is arithmetic. $\displaystyle \frac{1}{6},-\frac{1}{6},-\frac{1}{2},-\frac{5}{6},...$
If the sequence is arithmetic, there is a common difference between terms. $t_{n}-t_{n-1}=\displaystyle \left(\frac{1}{2}-\frac{1}{3}n\right)-\left(\frac{1}{2}-\frac{1}{3}(n-1)\right)$ $=\displaystyle \left(\frac{1}{2}-\frac{1}{3}n\right)-\left(\frac{1}{2}-\frac{1}{3}n+\frac{1}{3}\right)$ $=\displaystyle \frac{1}{2}-\frac{1}{3}n-\frac{1}{2}+\frac{1}{3}n-\frac{1}{3}$ $=-\displaystyle \frac{1}{3}$ $d=-\displaystyle \frac{1}{3}$, which is constant, so the sequence is arithmetic. $t_{1}=\displaystyle \frac{1}{2}-\frac{1}{3}(1)=\frac{3-2}{6}=\frac{1}{6}$ $t_{2}=\displaystyle \frac{1}{2}-\frac{1}{3}(2)=\frac{3-4}{6}=-\frac{1}{6}$ $t_{3}=\displaystyle \frac{1}{2}-\frac{1}{3}(3)=-\frac{1}{2}$ $t_{4}=\displaystyle \frac{1}{2}-\frac{1}{3}(4)=\frac{3-8}{6}=-\frac{5}{6}$