Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then $a_{6} \mathrm{a}_{16}$ is equal to :
Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then $a_{6} \mathrm{a}_{16}$ is equal to :
- [JEE MAIN 2021]
- A
$57$
- B
$72$
- C
$48$
- D
$36$
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