Answer: Option C
Let a be the first term and d be the common difference of an A.P. and
$$\eqalign{
& {S_n} = 2{n^2} + 5n \cr
& \therefore {S_1} = 2{\left( 1 \right)^2} + 5 \times 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 2 + 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 7 \cr
& \therefore {S_2} = 2{\left( 2 \right)^2} + 5 \times 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 + 10 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr
& \therefore {\text{First}}\,{\text{term}}\,\left( {{a_1}} \right) = 7\,{\text{and}} \cr
& {\text{Second}}\,{\text{term}}\,{a_2} = {S_2} - {S_1} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 - 7 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 \cr
& \therefore d = {a_2} - {d_1} \cr
& \,\,\,\,\,\,\,\,\,\,\, = 11 - 7 \cr
& \,\,\,\,\,\,\,\,\,\,\, = 4 \cr
& Now\,{a_n} = a + \left( {n - 1} \right)d \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + \left( {n - 1} \right)4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + 4n - 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4n + 3 \cr} $$