X marks the spot!

Let $d_1$ be the common difference of the arithmetic sequence $(a_x)$ . The first term is $a_1 = 1$ , so $a_x = 1 + (x - 1) d_1$ for all $n \geq 1$ . Since $a_2 > 1$ , the common difference $d_1$ is positive.

Similarly, let $d_2$ be the common difference of the arithmetic sequence $(b_x)$ . The first term is $b_1 = 1$ , so $b_x = 1 + (n - 1) d_2$ for all $n \geq 1$ . Since $b_2 > 1$ , the common difference $d_2$ is also positive.

We observe that $x - 1$ divides both $a_x - 1$ and $b_x - 1$ for all $x$ . Also, since $a_2 \leq b_2$ , $a_x \leq b_x$ for all $x$ .

If $a_x b_x = 2010$ , then $(a_x, b_x)$ must be one of the pairs $(2,1005), (3,670), (5,402), (6,335), (10,201), (15,134)$ , or $(30,67)$ . For each such pair, we compute the largest number dividing both $a_x - 1$ and $b_x - 1$ :

$\begin{array}{c|c|c} a_x & b_x & \gcd(a_x - 1, b_x - 1) \\ \hline 2 & 1005 & 1 \\ 3 & 670 & 1 \\ 5 & 402 & 1 \\ 6 & 335 & 1 \\ 10 & 201 & 1 \\ 15 & 134 & 7 \\ 30 & 67 & 1 \end{array}$

We see that the largest possible value of $x - 1$ is 7 (for $a_x = 15$ and $b_x = 134$ ), so the largest possible value of $x$ is $\boxed{8}$ . The corresponding arithmetic sequences are $a_x = 2x - 1$ and $b_x = 19x - 18$ .