Question

What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?

Answer 1

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ n=12\\ d=-10\\ a_{12}=13 \end{cases} \\\\\\ a_{12}=a_1+(12-1)d\implies 13=a_1+(12-1)(-10) \\\\\\ 13=a_1-110\implies \boxed{123=a_1}$

$\bf \\\\ -------------------------------\\\\ \textit{sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n}{2}(a_1+a_n)\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ a_n=\textit{value of the }n^{th}\ term\\ ------------\\ n=12\\ a_1=123\\ a_{12}=13 \end{cases} \\\\\\ S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)$

and surely you know how much that is.