(D) The sum of the first three terms of an arithmetic sequence is 36 and their product is 1140. Determine the three terms.

Let the first term of the arithmetic sequence be \( a \), and the common difference be \( d \). Therefore, the first three terms can be expressed as \( a \), \( a + d \), and \( a + 2d \). According to the problem: 1. The sum of the three terms is given by: \[ a + (a + d) + (a + 2d) = 3a + 3d = 36 \] This simplifies to: \[ a + d = 12 \quad (1) \] 2. The product of the three terms is given by: \[ a(a + d)(a + 2d) = 1140 \] Substituting \( d = 12 - a \) from equation (1) into the product gives us: \[ a(a + (12 - a))(a + 2(12 - a)) = 1140 \] Simplifying this: \[ a(12)(24 - a) = 1140 \] This expands to: \[ 12a(24 - a) = 1140 \] Dividing both sides by 12: \[ a(24 - a) = 95 \] This leads to the quadratic equation: \[ a^2 - 24a + 95 = 0 \] 3. To find \( a \), we can use the quadratic formula: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{24 \pm \sqrt{576 - 380}}{2} = \frac{24 \pm \sqrt{196}}{2} = \frac{24 \pm 14}{2} \] This yields two potential solutions for \( a \): \[ a = \frac{38}{2} = 19 \quad \text{or} \quad a = \frac{10}{2} = 5 \] 4. Now we can find \( d \) for both values: - If \( a = 19 \): \[ d = 12 - 19 = -7 \] The terms would then be \( 19, 12, 5 \). - If \( a = 5 \): \[ d = 12 - 5 = 7 \] The terms would then be \( 5, 12, 19 \). Thus, the three terms of the arithmetic sequence are \( 5, 12, 19 \).