$t_n = a + (n - 1)\,d$ — read aloud: t-n equals a, plus n minus one, times d. Here $a$ is the first term and $d$ the common difference. The $n - 1$ is the whole exam: the first term has taken zero steps, so by term $n$ only $n - 1$ steps have happened. Term 10 of 2, 5, 8, … is $2 + 9 \times 3 = 29$ — nine steps, not ten.

One fact, four sockets: solve it for $t_n$, for $a$, for $d$, or for $n$, and it answers four different exam questions. The ladder is the one you know — pick the formula, rearrange for the target, load the givens, then compute. And when the target is $n$, the answer must land on a whole number: term 11½ does not exist, and an answer like it means the arithmetic upstream slipped.