Determine whether the following sequence is an $A.P.$ or not. (Assume that the pattern continues.) If it is an $A.P.$,find its $n^{th}$ term: $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots$

(D) To check if the sequence $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots$ is an $A.P.$,we calculate the difference between consecutive terms.
Let $a_1 = \frac{1}{2}$,$a_2 = \frac{2}{3}$,$a_3 = \frac{3}{4}$,$a_4 = \frac{4}{5}$.
Calculate the common difference $d_1 = a_2 - a_1 = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.
Calculate the common difference $d_2 = a_3 - a_2 = \frac{3}{4} - \frac{2}{3} = \frac{9-8}{12} = \frac{1}{12}$.
Since $d_1 \neq d_2$,the difference between consecutive terms is not constant.
Therefore,the given sequence is not an $A.P.$