Question

Insira 4 meios aritméticos entre 3 e 27

Answer 1

✅ Após resolver os cálculos, concluímos que a progressão aritmética procurada é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\ P.A.(3,\:{\bf7.8,\, 12.6,\:17.4,\:22.2},\:27)\:\:\:}}\end{gathered} }$

Sejam os dados:

     $\Large\begin{cases} m = N\acute{u}mero\:de\:meios = 4\\n = m + 2 = 4 + 2 = 6\\A_{1} = Primeiro\:termo =3\\A_{6} = Sexto\:termo = 27\\r = Raz\tilde{a}o = \:?\end{cases}$

Para resolver esta questão devemos:

• Calcular a razão da P.A.:

             $\Large\displaystyle\text{ \begin{gathered}r = \frac{A_{n} - A_{1}}{n - 1} \end{gathered} }$

                 $\Large\displaystyle\begin{gathered} = \frac{27 - 3}{6 - 1}\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \frac{24}{5}\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = 4,8\end{gathered}$

           $\Large\displaystyle\begin{gathered} \therefore\:\:\:r = 4,8\end{gathered}$

• Calcular cada um dos 6 termos da P.A:

        $\Large\displaystyle\begin{gathered} A_{1} = 3\end{gathered}$

        $\Large\displaystyle\begin{gathered} A_{2} = A_{1} + r = 3 + 4,8 = 7,8\end{gathered}$

        $\Large\displaystyle\begin{gathered} A_{3} = A_{2} + r = 7,8 + 4,8 = 12,6\end{gathered}$

        $\Large\displaystyle\begin{gathered} A_{4} = A_{3} + r = 12,6 + 4,8 = 17,4\end{gathered}$

        $\Large\displaystyle\begin{gathered} A_{5} = A_{4} + r = 17,4 + 4,8 = 22,2\end{gathered}$

        $\Large\displaystyle\begin{gathered} A_{6} = A_{5} + r = 22,2 + 4,8 = 27\end{gathered}$

✅ Portanto, a progressão aritmética:

         $\Large\displaystyle\begin{gathered} P.A.(3,\:7.8,\, 12.6,\:17.4,\:22.2,\:27)\end{gathered}$

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