This is an arithmetic progression with first term $a = -12$ and common difference $d = 3$. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, we set $54 = \frac{n}{2}[-24 + (n-1)3]$. Simplifying gives $n^2 - 9n - 36 = 0$. Factoring $(n-12)(n+3) = 0$ yields $n=12$. Other options fail this quadratic equation.