Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.
+26387 Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.
\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n-3n^2 \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n-3n^2 - [ 4(n-1)-3(n-1)^2 ] \\ t_n &=&4n-3n^2 - 4(n-1) + 3(n-1)^2 \\ t_n &=&4n-3n^2 - 4n + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3n^2-6n+3 \\ t_n &=& 4 -6n+3 \\ \mathbf{t_n} & \mathbf{=} & \mathbf{7 -6n} \\\\ t_1 &=& 7 - 6\cdot 1 \\ t_1 &=& 7-6\\ \mathbf{t_1} &\mathbf{=}& \mathbf{1}\\\\ d &=& t_n - t_{n-1} \\ d &=& 7-6n -[ 7-6(n-1)]\\ d &=& 7-6n - 7 + 6(n-1)\\ d &=&-6n + 6(n-1)\\ d &=&-6n + 6n -6\\ \mathbf{d} &\mathbf{=}&\mathbf{-6}\\ \end{array}\)
+26387 Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.
\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n-3n^2 \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n-3n^2 - [ 4(n-1)-3(n-1)^2 ] \\ t_n &=&4n-3n^2 - 4(n-1) + 3(n-1)^2 \\ t_n &=&4n-3n^2 - 4n + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3n^2-6n+3 \\ t_n &=& 4 -6n+3 \\ \mathbf{t_n} & \mathbf{=} & \mathbf{7 -6n} \\\\ t_1 &=& 7 - 6\cdot 1 \\ t_1 &=& 7-6\\ \mathbf{t_1} &\mathbf{=}& \mathbf{1}\\\\ d &=& t_n - t_{n-1} \\ d &=& 7-6n -[ 7-6(n-1)]\\ d &=& 7-6n - 7 + 6(n-1)\\ d &=&-6n + 6(n-1)\\ d &=&-6n + 6n -6\\ \mathbf{d} &\mathbf{=}&\mathbf{-6}\\ \end{array}\)
