Precalculus (10th Edition)

In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(\frac{1}{2}-\frac{1}{3}(n+1))-(\frac{1}{2}-\frac{1}{3}n)=(\frac{1}{6}-\frac{1}{3}n)-(\frac{1}{2}-\frac{1}{3}n)=-\frac{1}{3}$, thus it is an arithmetic sequence. $a_1=\frac{1}{6}$ $a_2=-\frac{1}{6}$ $a_3=-\frac{1}{2}$ $a_4=-\frac{5}{6}$
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(\frac{1}{2}-\frac{1}{3}(n+1))-(\frac{1}{2}-\frac{1}{3}n)=(\frac{1}{6}-\frac{1}{3}n)-(\frac{1}{2}-\frac{1}{3}n)=-\frac{1}{3}$, thus it is an arithmetic sequence. $a_1=\frac{1}{2}-\frac{1}{3}(1)=\frac{1}{6}$ $a_2=\frac{1}{2}-\frac{1}{3}(2)=-\frac{1}{6}$ $a_3=\frac{1}{2}-\frac{1}{3}(3)=-\frac{1}{2}$ $a_4=\frac{1}{2}-\frac{1}{3}(4)=-\frac{5}{6}$