Respuesta :
[tex]\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=12\\
d=-10\\
a_{12}=13
\end{cases}
\\\\\\
a_{12}=a_1+(12-1)d\implies 13=a_1+(12-1)(-10)
\\\\\\
13=a_1-110\implies \boxed{123=a_1}[/tex]
[tex]\bf \\\\ -------------------------------\\\\ \textit{sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n}{2}(a_1+a_n)\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ a_n=\textit{value of the }n^{th}\ term\\ ------------\\ n=12\\ a_1=123\\ a_{12}=13 \end{cases} \\\\\\ S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)[/tex]
and surely you know how much that is.
[tex]\bf \\\\ -------------------------------\\\\ \textit{sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n}{2}(a_1+a_n)\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ a_n=\textit{value of the }n^{th}\ term\\ ------------\\ n=12\\ a_1=123\\ a_{12}=13 \end{cases} \\\\\\ S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)[/tex]
and surely you know how much that is.
