Official solution
\[\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } \]
\[ = {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)} \]
\[ \Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}\] (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)