Let a _ 1 , a _ 2 , , a _ 21 be an A . P . such that ∑ _ n = 1 ^ 20...

$\mathop{∑}_{n = 1}^{20}\dfrac{1}{{a}_{n}{a}_{n + 1}} = \dfrac{4}{9}$ $\text{[let common difference of}\leftAPisd = {a}_{n + 1} - {a}_{n}]$

$= \dfrac{1}{d}\mathop{∑}_{n = 1}^{20}\left(\dfrac{1}{{a}_{n}} - \dfrac{1}{{a}_{n + 1}}\right) ⇒ \dfrac{1}{d}\left[\left(\dfrac{1}{{a}_{1}} - \dfrac{1}{{a}_{2}}\right) + \left(\dfrac{1}{{a}_{2}} - \dfrac{1}{{a}_{3}}\right) + \ldots . \left(\dfrac{1}{{a}_{20}} - \dfrac{1}{{a}_{21}}\right)\right] = \dfrac{4}{9}$

$⇒ \dfrac{1}{d}\left(\dfrac{1}{{a}_{1}} - \dfrac{1}{{a}_{21}}\right) = \dfrac{4}{9}$

$⇒ \dfrac{1}{d}\left(\dfrac{{a}_{21} - {a}_{1}}{{a}_{1} \cdot {a}_{21}}\right) = \dfrac{4}{9} ⇒ \dfrac{20d}{d\left({a}_{1}{a}_{21}\right)} = \dfrac{4}{9} ⇒ {a}_{1}{a}_{21} = 45$

${a}_{1}\left({a}_{1} + 20d\right) = 45 . . . ( 1 )$

Now sum of first $21$ terms$= \dfrac{21}{2}\left(2{a}_{1} + 20d\right) = 189 ⇒ {a}_{1} + 10d = 9 . . . ( 2 )$

by using equation $( 1 )$ and $( 2 )$ we get ${a}_{1} = 3, d = \dfrac{3}{5}$ otherwise ${a}_{1} = 15, d = - \dfrac{3}{5}$

So, ${a}_{6}{a}_{16} = \left({a}_{1} + 5d\right)\left({a}_{1 +} + 15d\right) = 72$