O 78º termo da sua progressão aritmética é igual a: [tex]\large\displaystyle\text{$\begin{gathered}a_{78}=490\end{gathered}$}[/tex].
Para resolver sua questão, temos que utilizar a fórmula geral da P.A, dada da seguinte forma:
[tex]\Large\displaystyle\text{$\begin{gathered} \underline{\boxed{a_n=a_1+(n-1)\cdot r}}\end{gathered}$}[/tex]
Onde:
[tex]\Large\displaystyle\text{$\begin{gathered} a_n=termo\ geral\ da\ p.a \ ;\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} a_1=primeiro\ termo\ da\ p.a \ ;\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} n=n^{\circ}\ de\ termo\ da\ p.a \ ;\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} r=raz\tilde{a}o\ da\ p.a \ .\end{gathered}$}[/tex]
Substituindo, temos que:
[tex]\Large\displaystyle\text{$\begin{gathered}a_n=a_1+(n-1)\cdot r\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}a_{78}=-126+(78-1)\cdot 8\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}a_{78}=-126+77\cdot 8\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}a_{78}=-126+616\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\therefore \green{\underline{\boxed{a_{78}=490}}}\ \ (\checkmark).\end{gathered}$}[/tex]
Veja mais sobre:
- brainly.com.br/tarefa/497062
