Three numbers in an AP have a sum of 33 and a product of 1155. Find the numbers.

Step 1. The sum. $3a = 33$, so $a = 11$.

Step 2. The product. $11 \times (121 - d^2) = 1155$, so $121 - d^2 = 105$.

Step 3. Solve. $d^2 = 16$, so $d = 4$ or $d = -4$.

Step 4. The numbers. $7, 11, 15$. Check: $7 + 11 + 15 = 33$ and $7 \times 11 \times 15 = 1155$.