Question

Quantos termos devemos tomar na PA (-7, -3...) a fim de que a soma valha 3 150

Answer 1

$Ol\acute{a}~~~\mathbb{LAURA} \\ \\ [tex]Sendo~a~f\acute{o}rmula~de~P.A~temos: ~~~~\boxed{a_{n}= a_{1}+(n-1).r} ~---\ \textgreater \ \textcircled{1}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{S_{n}= \frac{(a_{1}+a_{n})}{2}.n }~----\ \textgreater \ \textcircled{2} \\ Dados: \\ a_{1}=-7 \\ a_{n}=? \\ n=? \\ r=-3-(-7)=4 \\ S_{n}=3150 \\ Substituindo~dados~na~\textcircled{1}~temos:$

$a_{n}=-7+(n-1).4 \\ \\ \boxed{a_{n}=4n-4 }~~------\ \textgreater \ (I) \\ \\ Igual~substituindo~dados~na~\textcircled{2} ~temos: \\ 3150= \frac{(-7+a_{n})}{2}.n \\ \\ \boxed{6300=(-7+a_{n}).n }----(II) \\ \\ Substituindo~(I)~em~(II) \\ \\ 6300=(-7+4n-11).n$

$6300=(4n-18)n \\ \\ 6300=4 n^{2}-18n~~~---\ \textgreater \ simplificando~dividindo~por~(2) \\ \\ 3150=2 n^{2}-9n \\ \\ 2 n^{2}-9n-3150=0 ~~~---\ \textgreater \ usando~a~\underbrace{f\acute{o}rmula~de~baskara } \\ ~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\frac{-b~\pm~ \sqrt{ b^{2}-4.a.c } }{2.a}} \\ ~Temos:a=2 \\ ~~~~~~~~~~~~b=-9 \\~~~~~~~~~~~~ c=-3150$

$Substituindo~dados~na~f\acute{o}rmula~de~baskara~temos: \\ \\ \frac{9\pm \sqrt{ (-9)^{2} -4.2(-3150)} }{4} \\ \\ \frac{9\pm \sqrt{25281} }{4} \\ \\ \frac{9\pm159}{4} = \begin{cases}n'= \frac{9+159}{4} =41 \\ \\ n''= \frac{9-159}{4} =-37,5~~---\ \textgreater \ descartamos\end{cases} \\ \\ \\ Por~tanto~\boxed{n=41}~~--\ \textgreater \ termo~a_{41}~~ou~41~termos$

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