Respuesta :

[tex]\bf \textit{sum of an arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ ----------\\ S_n=795\\ a_1=102\\ a_n=57 \end{cases} \\\\\\ 795=\cfrac{n(102+57)}{2}\implies 1590=159n \\\\\\ \cfrac{1590}{159}=n\implies 10=n\\\\ -------------------------------[/tex]

so the nth term is really the 10th term, and we know that's 57, thus

[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=10\\ a_{10}=57\\ a_1=102 \end{cases} \\\\\\ 57=102+(10-1)d\implies 57=102+9d\implies -45=9d \\\\\\ \cfrac{-45}{9}=d\implies -5=d[/tex]

so, that's the common difference... .so you'd surely know what the 3rd term is, notice the first one is 102.

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