Question

Numa p.g A1= ¼ e A7=16 calcule a razão dessa P.G

Answer 1

 Numa p.g A1= ¼ e A7=16 calcule a razão dessa P.G


1.q^6 = 16 
4

q^6 = 64 ==> q^6 = 2^6 ==> q = 2

Answer 2

✅ Após ter terminado os cálculos, concluímos que a razão da referida progressão geométrica é:

             $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf r = 2\:\:\:}}\end{gathered} }$

Sejam os dados:

                $\Large\begin{cases} A_{1} = 1/4\\A_{7} = 16\end{cases}$

Sabendo que a fórmula do termo geral da P.G. é:

$\Large\displaystyle\begin{gathered} \bf(I)\end{gathered}$         $\Large\displaystyle\begin{gathered} A_{n} = A_{1}\cdot q^{n - 1}\end{gathered}$

Substituindo os valores na equação "I", temos:

            $\Large\displaystyle\begin{gathered} 16 = \bigg(\frac{1}{4}\bigg)\cdot q^{7 - 1}\end{gathered}$

            $\Large\displaystyle\begin{gathered} 16 = \bigg(\frac{1}{4}\bigg)\cdot q^6\end{gathered}$

           $\Large\displaystyle\text{ \begin{gathered} \frac{16}{\frac{1}{4}} = q^{6}\end{gathered} }$

     $\Large\displaystyle\begin{gathered} 16\cdot 4 = q^{6}\end{gathered}$

           $\Large\displaystyle\begin{gathered} 64 = q^{6}\end{gathered}$

             $\Large\displaystyle\begin{gathered} q = \sqrt[6]{64}\end{gathered}$

              $\Large\displaystyle\begin{gathered} q = 2\end{gathered}$

✅ Portanto, a razão procurada é:

                 $\Large\displaystyle\begin{gathered} r = 2\end{gathered}$

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