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}\\ ----------\\ a_1=2\\ a_4=11\\ n=4 \end{cases} \\\\\\ a_4=2+(4-1)d\implies 11=2+(4-1)d\implies 11=2+3d \\\\\\ 9=3d\implies \cfrac{9}{3}=d\implies \boxed{3=d}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{now, what's the 50th term anyway?} \\\\\\ 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}\\ ----------\\ a_1=2\\ n=50\\ d=3 \end{cases} \\\\\\ a_{50}=2+(50-1)3\implies a_{50}=2+147\implies a_{50}=149\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n}{2}(a_1+a_n)\quad \begin{cases} n=50\\ a_1=2\\ a_{50}=149 \end{cases}\implies S_{50}=\cfrac{50}{2}(2+149)[/tex]
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