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

$\sum _{n=1}^{20}\frac{1}{a_n{a}_{n+1}}=\frac{4}{9}$ $\text{ [let common difference of }\left.\mathrm{AP} \mathrm{is} d={\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_{\mathrm{n}}\right]$

$=\frac{1}{d}\sum _{n=1}^{20}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)\Rightarrow \frac{1}{d}\left[\left(\frac{1}{a_1}-\frac{1}{a_2}\right)+\left(\frac{1}{a_2}-\frac{1}{a_3}\right)+\dots .\left(\frac{1}{a_{20}}-\frac{1}{a_{21}}\right)\right]=\frac{4}{9}$

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

$\Rightarrow \frac{1}{d}\left(\frac{a_{21}-a_1}{a_1·a_{21}}\right)=\frac{4}{9}\Rightarrow \frac{20d}{d\left(a_1{a}_{21}\right)}=\frac{4}{9}\Rightarrow a_1{a}_{21}=45$

$a_1\left(a_1+20d\right)=45 ...\left(1\right)$

Now sum of first $21$ terms$=\frac{21}{2}\left(2a_1+20d\right)=189\Rightarrow a_1+10d=9 ...\left(2\right)$

by using equation $\left(1\right)$ and $\left(2\right)$ we get $a_1=3, d=\frac{3}{5}$ otherwise $a_1=15, d=-\frac{3}{5}$

So, $a_6{a}_{16}=\left(a_1+5d\right)\left(a_{1+}+15d\right)=72$