How many terms of the AP 45, 41, 37, 33, ... are positive?

$a=45,d=-4$, so $a_n=45-4(n-1)=49-4n$. For $a_n$ > 0 we need $4n<49$, i.e. $n<12.25$, so $n$ can be 1, 2, ..., 12 - that is 12 terms$.(a_{12}=1$ is positive while $a_{13}=-3$ is not.) Rounding 12.25 up to 13 counts a negative term, and 49 is the constant in the formula, not a count.