Answer
$d=\dfrac{1}{4}$, a constant; so the sequence is an arithmetic sequence.
The first four terms are:
$t_1=\dfrac{11}{12}$
$t_2= \dfrac{7}{6}$
$t_3=\dfrac{17}{12}$
$t_4= \dfrac{5}{3}$
Work Step by Step
We know that a sequence is arithmetic when the common difference $d$ between the terms is constant.
In our case, we have:
$d=t_n-t_{n-1}=(\dfrac{2}{3}+\dfrac{1}{4}n)-(\dfrac{2}{3}-\dfrac{1}{4}(n-1))=\dfrac{2}{3}+\dfrac{n}{4}-\dfrac{2}{3}-\dfrac{n}{4}+\dfrac{1}{4}=\dfrac{1}{4}$
Thus, we see that $d=\dfrac{1}{4}$, which is a constant. So, the sequence is an arithmetic sequence.
Now, we find the first 4 terms:
$t_1=\dfrac{2}{3}+(1)(\dfrac{1}{4})= \dfrac{11}{12}$
$t_2=\dfrac{2}{3}+(2)(\dfrac{1}{4})= \dfrac{7}{6}$
$t_3=\dfrac{2}{3}+(3)(\dfrac{1}{4})= \dfrac{17}{12}$
$t_4=\dfrac{2}{3}+(4)(\dfrac{1}{4})= \dfrac{5}{3}$
