The 9th term of an AP is −32 and the sum of its 11th and 13th terms is −94.

Let a be the first term and d be the common difference of the AP. Then,

$a_{9}=-32$

$\Rightarrow a+(9-1) d=-32 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+8 d=-32 \quad \ldots \ldots(1)$

Now,

$a_{11}+a_{13}=-94$         (Given)

$\Rightarrow(a+10 d)+(a+12 d)=-94$

$\Rightarrow 2 a+22 d=-94$

$\Rightarrow a+11 d=-47 \quad \ldots(2)$

From (1) and (2), we get

$-32-8 d+11 d=-47$

$\Rightarrow 3 d=-47+32=-15$

$\Rightarrow d=-5$

Hence, the common difference of the AP is −5.