Are $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$ in an $AP$? If they form an $AP$,find the common difference $d$ and write three more terms.

(N/A) The given sequence is $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1)$
$a_{3} - a_{2} = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} = \sqrt{3}(\sqrt{3} - \sqrt{2})$
$a_{4} - a_{3} = \sqrt{12} - \sqrt{9} = 2\sqrt{3} - 3 = \sqrt{3}(2 - \sqrt{3})$
Since the difference $a_{k+1} - a_{k}$ is not constant for all $k$,the given sequence does not form an $AP$.