In an AP, the first term is −4, the last term is 29 and the sum of all its terms is 150.

Suppose there are n terms in the AP.

Here, a = −4, l = 29 and Sn = 150

$S_{n}=150$

$\Rightarrow \frac{n}{2}(-4+29)=150 \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

$\Rightarrow n=\frac{150 \times 2}{25}=12$

Thus, the AP contains 12 terms.

Let d be the common difference of the AP.

$\therefore a_{12}=29$

$\Rightarrow-4+(12-1) \times d=29 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 11 d=29+4=33$

$\Rightarrow d=3$

Hence, the common difference of the AP is 3.