Explanation

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