In an AP the first term is 2, the last term is 29 and sum of all the terms is 155.

Here, a = 2, l = 29 and Sn = 155
Let d be the common difference of the given AP and n be the total number of terms. 
Then Tn = 29
⇒ a + (n - 1)d = 29  
⇒ 2 + (n - 1)d = 29​                ...(i)

The sum of n terms of an AP is given by

$S_{n}=\frac{n}{2}[a+l]=155$

$\Rightarrow \frac{n}{2}[2+29]=\left(\frac{n}{2}\right) \times 31=155$

$\Rightarrow n=10$

Putting the value of n in (i), we get:
⇒ 2 + 9d = 29
⇒ 9d = 27
⇒ d = 3
Thus, the common difference of the given AP is 3.