$$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } $$

$$ = {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)} $$

$$ \Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}$$ (Given)

$$ \Rightarrow {1 \over d}\left( {{{{a_{21}} - {a_1}} \over {{a_1}{a_{21}}}}} \right) = {4 \over 9}$$

$$ \Rightarrow {1 \over d}\left( {{{{a_1} + 20d - {a_1}} \over {{a_1}{a_2}}}} \right) = {4 \over 9} \Rightarrow {a_1}{a_2} = 45$$ .... (1)

Now sum of first 21 terms = $${{21} \over 2}(2{a_1} + 20d) = 189$$

$$\Rightarrow$$ a₁ + 10d = 9 ..... (2)

For equation (1) & (2) we get

a₁ = 3 & d = $${3 \over 5}$$

or a₁ = 15 & d = $$ - {3 \over 5}$$

So, a₆ . a₁₆ = (a₁ + 5d) (a₁ + 15d)

$$\Rightarrow$$ a₆a₁₆ = 72

Option (b)