Answer
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(\frac{2}{3}+\frac{1}{4}(n+1))-(\frac{2}{3}+\frac{1}{4}n)=(\frac{11}{12}+\frac{1}{4}n)-(\frac{2}{3}+\frac{1}{4}n)=\frac{1}{4}$, thus it is an arithmetic sequence.
$a_1=\frac{11}{12}$
$a_2=\frac{7}{6}$
$a_3=\frac{17}{12}$
$a_4=\frac{5}{3}$
Work Step by Step
In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant.
Hence here: $a_{n+1}-a_n=(\frac{2}{3}+\frac{1}{4}(n+1))-(\frac{2}{3}+\frac{1}{4}n)=(\frac{11}{12}+\frac{1}{4}n)-(\frac{2}{3}+\frac{1}{4}n)=\frac{1}{4}$, thus it is an arithmetic sequence.
$a_1=\frac{2}{3}+\frac{1}{4}(1)=\frac{11}{12}$
$a_2=\frac{2}{3}+\frac{1}{4}(2)=\frac{7}{6}$
$a_3=\frac{2}{3}+\frac{1}{4}(3)=\frac{17}{12}$
$a_4=\frac{2}{3}+\frac{1}{4}(4)=\frac{5}{3}$
