Verify that each of the following is an $AP$,and then write its next three terms.
$5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

(A) Given sequence: $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$
Here,$a_{1}=5, a_{2}=\frac{14}{3}, a_{3}=\frac{13}{3}$ and $a_{4}=4$.
Calculate the common differences:
$a_{2}-a_{1} = \frac{14}{3} - 5 = \frac{14-15}{3} = -\frac{1}{3}$
$a_{3}-a_{2} = \frac{13}{3} - \frac{14}{3} = -\frac{1}{3}$
$a_{4}-a_{3} = 4 - \frac{13}{3} = \frac{12-13}{3} = -\frac{1}{3}$
Since the difference between consecutive terms is constant $(d = -\frac{1}{3})$,the given sequence is an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = 4 + (-\frac{1}{3}) = \frac{12-1}{3} = \frac{11}{3}$
$a_{6} = a_{5} + d = \frac{11}{3} + (-\frac{1}{3}) = \frac{10}{3}$
$a_{7} = a_{6} + d = \frac{10}{3} + (-\frac{1}{3}) = \frac{9}{3} = 3$
Thus,the next three terms are $\frac{11}{3}, \frac{10}{3}, 3$.