The 11 th term of an AP -5,

Given AP, $5, \frac{-5}{2}, 0, \frac{5}{2}$

Here, $a=-5, d=\frac{-5}{2}+5=\frac{5}{2}$

$\therefore \quad a_{11}=a+(11-1) d \quad\left[\because a_{n}=a+(n-1) d\right]$

$=-5+(10) \times \frac{5}{2}$

$=-5+25=20$