3 Determine the first three terms and the \( n^{\text {th }} \) term of the sequence if \( 3.1 \mathrm{~S}_{n}=43 n-6 n^{2} \) \( 3.2 \mathrm{~S}_{n}=n+2 n^{2} \) \( 3.3 \quad \mathrm{~S}_{n}=\frac{31}{2} n-\frac{3}{2} n^{2} \)

To find the first three terms and the \( n^{\text{th}} \) term of each sequence, we can evaluate the given sums \( S_n \) at different values of \( n \) (specifically 1, 2, and 3) and then derive the individual terms. For \( S_n = 43n - 6n^2 \): - \( S_1 = 43(1) - 6(1)^2 = 43 - 6 = 37 \) - \( S_2 = 43(2) - 6(2)^2 = 86 - 24 = 62 \) - \( S_3 = 43(3) - 6(3)^2 = 129 - 54 = 75 \) Thus, the first three terms are 37, 62, 75, and the \( n^{\text{th}} \) term \( T_n = S_n - S_{n-1} \) can be calculated as \( 43n - 6n^2 - (43(n-1) - 6(n-1)^2) \). For \( S_n = n + 2n^2 \): - \( S_1 = 1 + 2(1)^2 = 1 + 2 = 3 \) - \( S_2 = 2 + 2(2)^2 = 2 + 8 = 10 \) - \( S_3 = 3 + 2(3)^2 = 3 + 18 = 21 \) The first three terms are 3, 10, 21, and the \( n^{\text{th}} \) term can be determined similarly. For \( S_n = \frac{31}{2} n - \frac{3}{2} n^2 \): - \( S_1 = \frac{31}{2}(1) - \frac{3}{2}(1)^2 = \frac{31}{2} - \frac{3}{2} = \frac{28}{2} = 14 \) - \( S_2 = \frac{31}{2}(2) - \frac{3}{2}(2)^2 = \frac{62}{2} - \frac{12}{2} = \frac{50}{2} = 25 \) - \( S_3 = \frac{31}{2}(3) - \frac{3}{2}(3)^2 = \frac{93}{2} - \frac{27}{2} = \frac{66}{2} = 33 \) Thus, the first three terms are 14, 25, 33, and the \( n^{\text{th}} \) term similarly can be analyzed. In summary: - \( S_n = 43n - 6n^2 \): First three terms: 37, 62, 75. - \( S_n = n + 2n^2 \): First three terms: 3, 10, 21. - \( S_n = \frac{31}{2} n - \frac{3}{2} n^2 \): First three terms: 14, 25, 33.