Respuesta :
[tex]\bf \begin{array}{ll}
n&a_n\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
6&50\\
7&50+d\\
8&50+d+d\\
9&50+d+d+d\\
10&50+d+d+d+d\\
11&50+d+d+d+d+d\\
&35
\end{array}
\\\\\\
50+5d=35\implies 5d=-15\implies d=\cfrac{-15}{5}\implies d=-3
\\\\\\
\textit{we know d = -3, and we know }a_{11}=35\qquad thus[/tex]
[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=-3\\ n=11\\ a_{11}=35 \end{cases} \\\\\\ a_{11}=a_1+(11-1)(-3)\implies 35=a_1+(11-1)(-3) \\\\\\ 35=a_1-30\implies 65=a_1[/tex]
now we know what "d" is, and what a₁ is, so let's check the 30th term,
[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=-3\\ n=30\\ a_{1}=65 \end{cases} \\\\\\ a_{30}=65+(30-1)(-3)\implies a_{30}=65-87\implies a_{30}=-22[/tex]
[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=-3\\ n=11\\ a_{11}=35 \end{cases} \\\\\\ a_{11}=a_1+(11-1)(-3)\implies 35=a_1+(11-1)(-3) \\\\\\ 35=a_1-30\implies 65=a_1[/tex]
now we know what "d" is, and what a₁ is, so let's check the 30th term,
[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=-3\\ n=30\\ a_{1}=65 \end{cases} \\\\\\ a_{30}=65+(30-1)(-3)\implies a_{30}=65-87\implies a_{30}=-22[/tex]
