Answer: Option D
$$\eqalign{
& {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\sqrt 7 ,\,\sqrt {28} ,\,\sqrt {63} ,\,...... \cr
& \Rightarrow \sqrt 7 ,\,\sqrt {4 \times 7} ,\,\sqrt {9 \times 7} ,\,..... \cr
& \Rightarrow \sqrt 7 ,\,2\sqrt 7 ,\,3\sqrt 7 ,...... \cr
& \therefore Here\,\,a = \sqrt 7 \,{\text{and}} \cr
& d = 2\sqrt 7 - \sqrt 7 = \sqrt 7 \cr
& \therefore {\text{Next}}\,{\text{term}} = 4\sqrt 7 \cr
& = \sqrt {\left( {16 \times 7} \right)} \cr
& = \sqrt {112} \cr} $$