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}\\ ----------\\ d=5\\ a_3=8\\ n=3 \end{cases} \\\\\\ a_3=a_1+(3-1)5\implies 8=a_1+(2)5\implies 8=a_1+10 \\\\\\ \boxed{-2=a_1} \\\\\\ a_n=-2+(n-1)5[/tex]

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