1. **State the problem:** We need to find five consecutive terms of an arithmetic sequence whose sum is 40 and the product of the first, middle, and last terms is 224. 2. **Define variables:** Let the five consecutive terms be $x - 2d, x - d, x, x + d, x + 2d$, where $x$ is the middle term and $d$ is the common difference. 3. **Sum of the terms:** The sum is given by $$ (x - 2d) + (x - d) + x + (x + d) + (x + 2d) = 5x $$ Since the sum is 40, we have $$ 5x = 40 $$ $$ \Rightarrow x = \frac{40}{5} = 8 $$ 4. **Product of first, middle, and last terms:** The product is $$ (x - 2d) \times x \times (x + 2d) = 224 $$ Substitute $x = 8$: $$ (8 - 2d) \times 8 \times (8 + 2d) = 224 $$ 5. **Simplify the product:** $$ 8 \times (8 - 2d)(8 + 2d) = 224 $$ Use difference of squares: $$ (8 - 2d)(8 + 2d) = 8^2 - (2d)^2 = 64 - 4d^2 $$ So, $$ 8 \times (64 - 4d^2) = 224 $$ 6. **Simplify further:** $$ 512 - 32d^2 = 224 $$ 7. **Solve for $d^2$:** $$ 512 - 224 = 32d^2 $$ $$ 288 = 32d^2 $$ $$ d^2 = \frac{288}{32} = 9 $$ 8. **Find $d$:** $$ d = \pm 3 $$ 9. **Find the terms:** For $d = 3$: $$ 8 - 2(3) = 8 - 6 = 2 $$ $$ 8 - 3 = 5 $$ $$ 8 $$ $$ 8 + 3 = 11 $$ $$ 8 + 2(3) = 8 + 6 = 14 $$ For $d = -3$, the terms are reversed but the same set. **Final answer:** The five consecutive terms are $2, 5, 8, 11, 14$.