Since, the first 11 terms in AP
$d=2$
$\therefore$ $a_{11}=a+10d=a+20$
The middle term of AP is $T_6=a+5d=a+10$
For the next 11 terms in GP, $r=2$
$\therefore$ The middle term of GP is $b{\left(2\right)}^5$ where b is the first term of a GP which is the last term of AP.
$\therefore$ $b{\left(2\right)}^5=\left(a+20\right) 32$
According to the given condition,
$a+10=\left(a+20\right) 32$
$\Rightarrow$ $31a=10-640 \Rightarrow a= -\frac{630}{31}$
$\therefore$ Middle term of entire sequence is ${11}^{th}$ term.
$\therefore$ $T_{11}= -\frac{630}{31}+10\times d$
$= -\frac{630}{31}+10 \times 2= -\frac{10}{31}$