## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$d=\dfrac{-1}{3}$; the sequence is an arithmetic sequence. The first four terms are: $t_1=\dfrac{1}{2}-\dfrac{1}{3}(1)= \dfrac{1}{6}$ $t_2=\dfrac{1}{2}-\dfrac{1}{3}(2)= -\dfrac{1}{6}$ $t_3=\dfrac{1}{2}-\dfrac{1}{3}(3)= \dfrac{-1}{2}$ $t_4=\dfrac{1}{2}-\dfrac{1}{3}(4)= -\dfrac{5}{6}$
The sequence is known to be arithmetic when there is a common difference $d$ between the terms. Now, $d=t_n-t_{n-1}=(\dfrac{1}{2}-\dfrac{1}{3}n)-(\dfrac{1}{2}-\dfrac{1}{3}(n-1))=\dfrac{1}{2}-\dfrac{n}{3}-\dfrac{1}{2}+\dfrac{n}{3}-\dfrac{1}{3}=\dfrac{-1}{3}$ Therefore, the sequence is an arithmetic sequence. Now, $t_1=\dfrac{1}{2}-\dfrac{1}{3}(1)= \dfrac{1}{6}$ $t_2=\dfrac{1}{2}-\dfrac{1}{3}(2)= -\dfrac{1}{6}$ $t_3=\dfrac{1}{2}-\dfrac{1}{3}(3)= \dfrac{-1}{2}$ $t_4=\dfrac{1}{2}-\dfrac{1}{3}(4)= -\dfrac{5}{6}$