In an AP the first term is 22, nth term is −11 and sum to first nth terms is 66.

Here, a = 22, Tn = -11 and Sn = 66
Let d be the common difference of the given AP. 
Then Tn = -11 
⇒ a + (n - 1)d = 22 + (n - 1)d = -11
⇒ (n - 1)d = -33        ...(i)

The sum of n terms of an AP is given by

$S_{n}=\frac{n}{2}[2 a+(n-1) d]=66$                    [Substituting the value of $(n-1) d$ from (i)]

$\Rightarrow \frac{n}{2}[2 \times 22+(-33)]=\left(\frac{n}{2}\right) \times 11=66$

$\Rightarrow n=12$

Putting the value of n in (i), we get:
11d = -33
⇒ d = -3
Thus, n = 12 and d = -3