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

To determine the first three terms and the \( n^{\text{th}} \) term of a sequence given its partial sum \( S_n \), we'll use the following approach: 1. **First Three Terms (\( a_1, a_2, a_3 \))**: - \( a_1 = S_1 \) - \( a_2 = S_2 - S_1 \) - \( a_3 = S_3 - S_2 \) 2. **\( n^{\text{th}} \) Term (\( a_n \))**: - \( a_n = S_n - S_{n-1} \) Let's apply this method to both problems: --- ## Problem 3.1 **Given:** \[ S_n = 43n - 6n^2 \] ### Finding the First Three Terms 1. **First Term (\( a_1 \)):** \[ a_1 = S_1 = 43(1) - 6(1)^2 = 43 - 6 = 37 \] 2. **Second Term (\( a_2 \)):** \[ S_2 = 43(2) - 6(2)^2 = 86 - 24 = 62 \] \[ a_2 = S_2 - S_1 = 62 - 37 = 25 \] 3. **Third Term (\( a_3 \)):** \[ S_3 = 43(3) - 6(3)^2 = 129 - 54 = 75 \] \[ a_3 = S_3 - S_2 = 75 - 62 = 13 \] **First Three Terms:** \[ a_1 = 37,\quad a_2 = 25,\quad a_3 = 13 \] ### Finding the \( n^{\text{th}} \) Term \[ a_n = S_n - S_{n-1} = [43n - 6n^2] - [43(n-1) - 6(n-1)^2] \] \[ = 43n - 6n^2 - 43n + 43 - 6(n^2 - 2n + 1) \] \[ = 43n - 6n^2 - 43n + 43 - 6n^2 + 12n - 6 \] \[ = (-12n) + 37 \] \[ \boxed{a_n = 49 - 12n} \] --- ## Problem 3.3 **Given:** \[ S_n = \frac{31}{2}n - \frac{3}{2}n^2 \] ### Finding the First Three Terms 1. **First Term (\( a_1 \)):** \[ a_1 = S_1 = \frac{31}{2}(1) - \frac{3}{2}(1)^2 = \frac{31}{2} - \frac{3}{2} = \frac{28}{2} = 14 \] 2. **Second Term (\( a_2 \)):** \[ S_2 = \frac{31}{2}(2) - \frac{3}{2}(2)^2 = 31 - 6 = 25 \] \[ a_2 = S_2 - S_1 = 25 - 14 = 11 \] 3. **Third Term (\( a_3 \)):** \[ S_3 = \frac{31}{2}(3) - \frac{3}{2}(3)^2 = \frac{93}{2} - \frac{27}{2} = \frac{66}{2} = 33 \] \[ a_3 = S_3 - S_2 = 33 - 25 = 8 \] **First Three Terms:** \[ a_1 = 14,\quad a_2 = 11,\quad a_3 = 8 \] ### Finding the \( n^{\text{th}} \) Term \[ a_n = S_n - S_{n-1} = \left(\frac{31}{2}n - \frac{3}{2}n^2\right) - \left(\frac{31}{2}(n-1) - \frac{3}{2}(n-1)^2\right) \] \[ = \frac{31}{2}n - \frac{3}{2}n^2 - \frac{31}{2}n + \frac{31}{2} - \frac{3}{2}(n^2 - 2n + 1) \] \[ = \frac{31}{2}n - \frac{3}{2}n^2 - \frac{31}{2}n + \frac{31}{2} - \frac{3}{2}n^2 + 3n - \frac{3}{2} \] \[ = ( -3n ) + 17 \] \[ \boxed{a_n = 17 - 3n} \] --- ## Summary - **Problem 3.1:** - **First Three Terms:** \( 37, \ 25, \ 13 \) - **\( n^{\text{th}} \) Term:** \( a_n = 49 - 12n \) - **Problem 3.3:** - **First Three Terms:** \( 14, \ 11, \ 8 \) - **\( n^{\text{th}} \) Term:** \( a_n = 17 - 3n \)