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

Let's denote the first term of the arithmetic sequence as \( a \) and the common difference as \( d \). The three terms can be expressed as \( a \), \( a + d \), and \( a + 2d \). From the problem, we have the following equations: 1. The sum of the three terms: \[ a + (a + d) + (a + 2d) = 36 \] Simplifying this, we get: \[ 3a + 3d = 36 \quad \Rightarrow \quad a + d = 12 \quad \text{(Equation 1)} \] 2. The product of the three terms: \[ a \cdot (a + d) \cdot (a + 2d) = 1140 \] Substituting \( a + d \) with \( 12 \) (from Equation 1), we rewrite the product as: \[ a \cdot 12 \cdot (12 + d) = 1140 \] Simplifying, we get: \[ 12a(12 + d) = 1140 \quad \Rightarrow \quad a(12 + d) = 95 \quad \text{(Equation 2)} \] Now, from Equation 1, we can express \( d \) in terms of \( a \): \[ d = 12 - a \] Substituting this into Equation 2 gives us: \[ a(12 + (12 - a)) = 95 \] which simplifies to: \[ a(24 - a) = 95 \] Expanding this, we have: \[ 24a - a^2 = 95 \quad \Rightarrow \quad a^2 - 24a + 95 = 0 \] Now, we can use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 1 \cdot 95}}{2 \cdot 1} \] Calculating the discriminant: \[ 576 - 380 = 196 \] So: \[ a = \frac{24 \pm 14}{2} \] This gives us two potential solutions: 1. \( a = \frac{38}{2} = 19 \) 2. \( a = \frac{10}{2} = 5 \) Now substituting back to find \( d \): 1. If \( a = 19 \): \[ d = 12 - 19 = -7 \] The sequence becomes \( 19, 12, 5 \). 2. If \( a = 5 \): \[ d = 12 - 5 = 7 \] The sequence becomes \( 5, 12, 19 \). Both sequences are valid, so the three terms of the arithmetic sequence are \( 5, 12, 19 \) (or alternatively, if ordered differently, \( 19, 12, 5 \)).