✅ Resolução de Progressão Geométrica (P.G.):
Pra resolver P.G, devemos utilizar a seguinte fórmula:
[tex]\large\displaystyle\text{$\begin{gathered}A_{n} = A_{1}.q^{n - 1} \end{gathered}$}[/tex]
Onde:
[tex]\large\displaystyle\text{$\begin{gathered}A_{n} = Termo\:procurado \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}A_{1} = Primeiro\:termo \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}q = Raz\tilde{a}o \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}n = Ordem\:termo\:procurado \end{gathered}$}[/tex]
Se:
[tex]\large\displaystyle\text{$\begin{gathered}A_{1} = 32 \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}A_{n} = 1024 \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}q = 2 \end{gathered}$}[/tex]
Então:
[tex]\large\displaystyle\text{$\begin{gathered}1024 = 32.2^{n - 1} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\frac{1024}{32} = 2^{n -1} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}32 = 2^{n - 1} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}32 = \frac{2^{n} }{2^{1} } \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}2.32 = 2^{n} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}64 = 2^{n} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\!\diagup\!\!\!\!2^{6} = \!\diagup\!\!\!2^{n} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}6 = n \end{gathered}$}[/tex]
✅ Portanto, o número de termos da P.G é:
[tex]\large\displaystyle\text{$\begin{gathered}n = 6 \end{gathered}$}[/tex]
✅ Portanto, a resposta correta é a letra B.
Saiba mais:
