How many terms of the AP 30, 27, 24, … must be added to give 156? Explain why there are two answers.

Step 1. The equation. $n(63 - 3n) = 312$, so $n^2 - 21n + 104 = 0$.

Step 2. Solve. $(n - 8)(n - 13) = 0$, so $n = 8$ or $n = 13$.

Step 3. Why two. Terms $9$ to $13$ are $6, 3, 0, -3, -6$. They add up to $0$, so the sum stays at $156$.