The sequence of four numbers $a_1$, $a_2$, $a_3$, and $a_4$ is such that each number after the first is $a_1-1$ greater than the preceding number. What is the value of $a_1$? (1) $a_2=15$ (2) $a_4=29$

Wait, the common difference is a1 - 1? That's unusual, but I guess it means d depends on the first term. Took me a second to set up the equations.

Got a2 = a1 + (a1 - 1) = 2a1 - 1. Statement 1 gives 2a1 - 1 = 15, so that works. Statement 2 gives a4 = a1 + 3(a1 - 1) = 29, also solvable. So D?

Is this really 500-level? The a1-1 as the difference tripped me up. I kept thinking it was just a constant like usual.

Quick note: since the difference contains a1 itself, you can't just treat it like a normal arithmetic sequence with a fixed d. Set up each term in terms of a1 and it becomes straightforward.