Answer
$d=\displaystyle \frac{1}{4}$, which is constant, so the sequence is arithmetic.
$\displaystyle \frac{11}{12},\frac{7}{6},\frac{17}{12},\frac{5}{3},...$
Work Step by Step
If the sequence is arithmetic, there is a common difference between terms.
$t_{n}-t_{n-1}=\displaystyle \left(\frac{2}{3}+\frac{1}{4}n\right)-\left(\frac{2}{3}+\frac{1}{4}(n-1)\right)$
$=\displaystyle \left(\frac{2}{3}+\frac{1}{4}n\right)-\left(\frac{2}{3}+\frac{1}{4}n-\frac{1}{4}\right)$
$=\displaystyle \frac{2}{3}+\frac{1}{4}n-\frac{2}{3}-\frac{1}{4}n+\frac{1}{4}$
$=\displaystyle \frac{1}{4}$
$d=\displaystyle \frac{1}{4}$, which is constant, so the sequence is arithmetic.
$t_{1}=\displaystyle \frac{2}{3}+\frac{1}{4}(1)=\frac{8+3}{12}=\frac{11}{12}$
$t_{2}=\displaystyle \frac{11}{12}+\frac{1}{4}=\frac{11+3}{12}=\frac{14}{12}=\frac{7}{6}$
$t_{3}=\displaystyle \frac{7}{6}+\frac{1}{4}=\frac{14+3}{12}=\frac{17}{12}$
$t_{4}=\displaystyle \frac{17}{12}+\frac{1}{4}=\frac{17+3}{12}=\frac{20}{12}=\frac{5}{3}$
