Answer: Option D
$$\eqalign{
& {a_n} = a + \left( {n - 1} \right)d \cr
& {a_9} = 449 \cr
& \,\,\,\,\,\, = a + \left( {9 - 1} \right)d \cr
& \,\,\,\,\,\, = a + 8d\,.....\left( 1 \right) \cr
& {a_{449}} = 9 \cr
& \,\,\,\,\,\,\,\,\, = a + \left( {449 - 1} \right)d \cr
& \,\,\,\,\,\,\,\,\, = a + 448d\,.....\left( 2 \right) \cr
& {\text{Subtracting}} \cr
& 440d = - 440 \cr
& \Rightarrow d = \frac{{ - 440}}{{440}} = - 1 \cr
& {\text{and}}\,a + 8d = 449 \cr
& \Rightarrow a \times 8 \times \left( { - 1} \right) = 449 \cr
& \Rightarrow a = 449 + 8 = 457 \cr
& \therefore 0 = a + \left( {n - 1} \right)d \cr
& \Rightarrow 0 = 457 + \left( {n - 1} \right)\left( { - 1} \right) \cr
& \Rightarrow 0 = 457 - n + 1 \cr
& \Rightarrow n = 458 \cr
& \therefore {458^{{\text{th}}}}\,{\text{term}} = 0 \cr} $$