Given that (2x+1), (7x+1) and (10x+1) are the first three consecutive terms of an AP. Find the value of x and the common difference.

Let the terms be $a_1, a_2, a_3$.

$$a_1 = 2x+1$$ $$a_2 = 7x+1$$ $$a_3 = 10x+1$$

The common difference $d$ is $a_2 - a_1$ and $a_3 - a_2$. Therefore, $a_2 - a_1 = a_3 - a_2$.

$$7x+1-2x-1 = 10x+1-7x-1$$ $$5x = 3x$$ $$5x - 3x = 0$$ $$2x = 0$$ $$x = 0$$

Substitute $x=0$ into the terms:
$a_1 = 2(0)+1 = 1$
$a_2 = 7(0)+1 = 1$
$a_3 = 10(0)+1 = 1$
The common difference $d = a_2 - a_1 = 1 - 1 = 0$.

The value of $x$ is $\boxed{0}$ and the common difference is $\boxed{0}$.