Answer & Solution
Answer: Option A
Solution:
$$\eqalign{
& {T_3} = a + 2d = 13\,.....\,\left( 1 \right) \cr
& {T_5} = a + 4d = 21\,.....\,\left( 2 \right) \cr
& {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr
& d = 4\& a = 5 \cr
& {T_{13}} = a + 12d \cr
& \,\,\,\,\,\,\,\,\, = 5 + 12\left( 4 \right) \cr
& \,\,\,\,\,\,\,\,\, = 5 + 48 \cr
& \,\,\,\,\,\,\,\,\, = 53 \cr} $$