Respuesta :
[tex]\bf 7~~,~~\stackrel{7+6}{13}......601\qquad \qquad \stackrel{\textit{common difference}}{d = 6} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=7\\ d=6\\ a_n=601 \end{cases} \\\\\\ 601=7+(n-1)6\implies 601=7+6n-6\implies 601=1+6n \\\\\\ 600=6n\implies \cfrac{600}{6}=n\implies 100=n \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ \cline{1-1} a_1=7\\ a_n=601\\ n=100 \end{cases}\implies S_{100}=\cfrac{100(7+601)}{2} \\\\\\ S_{100}=\cfrac{60800}{2}\implies S_{100}=30400[/tex]