Are $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$ in an $AP$? If they form an $AP$,find the common difference $d$ and write the next three terms.

(A) The given sequence is $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
We can simplify the terms as:
$a_1 = \sqrt{2} = 1\sqrt{2}$
$a_2 = \sqrt{8} = 2\sqrt{2}$
$a_3 = \sqrt{18} = 3\sqrt{2}$
$a_4 = \sqrt{32} = 4\sqrt{2}$
To check if it is an $AP$,we find the common difference $d = a_{k+1} - a_k$:
$a_2 - a_1 = 2\sqrt{2} - 1\sqrt{2} = \sqrt{2}$
$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
$a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$
Since the difference $d = \sqrt{2}$ is constant,the sequence is an $AP$.
The next three terms are:
$a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}$
$a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$
$a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$