Are $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$ in $AP$? If they form an $AP$,find the common difference $d$ and write the next three terms.

(A) The given sequence is $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$
To check if the sequence is in $AP$,we calculate the difference between consecutive terms:
$a_2 - a_1 = (3 + \sqrt{2}) - 3 = \sqrt{2}$
$a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2}$
$a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2}$
Since the difference $a_{k+1} - a_k$ is constant,the sequence is an $AP$ with common difference $d = \sqrt{2}$.
The next three terms are:
$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$
$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$
$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$