The sum of the first three terms of an arithmetic sequence is 36 and their product is 1140 . Determine the three terms. A certain sequence has a constant second difference of 8 . The sixth term of the sequence is

To find the three terms of the arithmetic sequence, let's denote them as \(a - d\), \(a\), and \(a + d\). The sum of these three terms is given by: \[ (a - d) + a + (a + d) = 3a = 36 \implies a = 12 \] Now substituting \(a\) back, we find the individual terms sum to \(36\). For the product, we have: \[ (a - d) \cdot a \cdot (a + d) = 1140 \] This translates to: \[ (12 - d)(12)(12 + d) = 1140 \] Expanding this yields: \[ 144(144 - d^2) = 1140 \implies 144 - d^2 = 7.91667 \implies d^2 = 136.0833 \] Thus \(d = 11.67\) (approximately). Therefore, the terms are \(12 - 11.67\), \(12\), and \(12 + 11.67\) which leads to the numbers \(0.33\), \(12\), and \(23.67\). For the second part: A sequence with a constant second difference of 8 is quadratic and can be described as \(T_n = an^2 + bn + c\). Knowing the second difference, we determine \(a = 4\) due to the relationship \(2a = 8\). The value of the sixth term can be calculated as: \[ T_6 = 4(6^2) + b(6) + c = 144 + 6b + c \] Without the values of \(b\) and \(c\), this expression provides a generalized form. Given particular \(b\) and \(c\), you could solve for the precise term. Enjoy your math adventure!