Let a_{1},a_{2},,a_{21} be an AP such that \sum_{n = 1}^{20}\mspace{2mu}\frac{1}{a_{n}a_{n + 1}} = \frac{4}{9}. If the sum of this AP is 189 , then a_{6}a_{16} is equal to: Sequence & Series (Mathematics) | DivineJEE