Question

encontre a PA crescente de três termos sabendo que a soma desss termos é -1/2 e o produto dos extremos é -1/12​

Answer 1

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

        $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf P.A.\bigg(-\frac{1}{2},\,-\frac{1}{6},\,\frac{1}{6}\bigg)\:\:\:}}\end{gathered} }$

Determinando a progressão aritmética

Sabemos que os termos da P.A. são:

                      $\Large\begin{cases} A_{1} = x\\A_{2} = x + r\\A_{3} = x + 2r\end{cases}$

Montando o sistema de equações a partir dos dados fornecidos, temos:

         $\Large\begin{cases} x + x + r + x + 2r = -1/2\\x\cdot(x + 2r) = -1/12\end{cases}$

Simplificando o referido sistema, temos:

            $\Large\begin{cases} 3x + 3r = -1/2\:\:\:\:\:\:\:\:\:\bf I\\x^{2} + 2xr = -1/12\:\:\:\:\bf II\end{cases}$

Isolando "x" na equação "I", temos:

                  $\Large\displaystyle\begin{gathered} 3x + 3r = -\frac{1}{2}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} 3(x + r) = -\frac{1}{2}\end{gathered}$

                       $\Large\displaystyle\text{ \begin{gathered} x + r = \frac{-\dfrac{1}{2}}{3}\end{gathered} }$

                       $\Large\displaystyle\begin{gathered} x + r = -\frac{1}{2}\cdot\frac{1}{3}\end{gathered}$

                       $\Large\displaystyle\begin{gathered} x + r = -\frac{1}{6}\end{gathered}$

                                $\Large\displaystyle\begin{gathered} x = -\frac{1}{6} - r\end{gathered}$

                                $\Large\displaystyle\begin{gathered} x = \frac{-1 - 6r}{6}\end{gathered}$

                             

Substituindo o valor de "x" na equação "II", temos:

        $\Large\displaystyle\begin{gathered} \bigg(\frac{-1 - 6r}{6}\bigg)^{2} + 2\cdot\bigg(-\frac{1}{6} - r\bigg)\cdot r = -\frac{1}{12}\end{gathered}$

                $\Large\displaystyle\text{ \begin{gathered} \frac{(-1 - 6r)^{2}}{6^{2}} + 2r\cdot\bigg(-\frac{1}{6} - r\bigg) = -\frac{1}{12}\end{gathered} }$

                        $\Large\displaystyle\text{ \begin{gathered} \frac{1 + 12r + 36r^{2}}{36} - \frac{2r}{6} - 2r^{2} = -\frac{1}{12}\end{gathered} }$

                    $\Large\displaystyle\text{ \begin{gathered} \frac{1 + 12r + 36r^{2} - 12r - 72r^{2}}{36} = -\frac{1}{12}\end{gathered} }$

          $\Large\displaystyle\begin{gathered} 12 + 144r + 432r^{2} - 144r - 864r^{2} = -36\end{gathered}$

$\Large\displaystyle\begin{gathered} 12 + 144r + 432r^{2} - 144r - 864r^{2} + 36 = 0\end{gathered}$

        $\Large\displaystyle\begin{gathered} -432r^{2} + 48 = 0\end{gathered}$

               $\Large\displaystyle\begin{gathered} -9r^{2} + 1 = 0\end{gathered}$

                        $\Large\displaystyle\begin{gathered} -9r^{2} = -1\end{gathered}$

                           $\Large\displaystyle\begin{gathered} 9r^{2} = 1\end{gathered}$

                              $\Large\displaystyle\begin{gathered} r^{2} = \frac{1}{9}\end{gathered}$

                              $\Large\displaystyle\text{ \begin{gathered} r = \pm\sqrt{\frac{1}{9}}\end{gathered} }$

                              $\Large\displaystyle\begin{gathered} r = \pm\frac{1}{3}\end{gathered}$

Como estamos querendo apenas uma P.A. crescente então devemos escolher a razão positiva, ou seja:

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

Calculando o valor de "x", temos:

                     $\Large\displaystyle\begin{gathered} x = -\frac{1}{6} - \frac{1}{3} = - \frac{1}{2}\end{gathered}$

Então, os valores dos três termos são:

      $\Large\displaystyle\begin{gathered} A_{1} = x = -\frac{1}{2}\end{gathered}$

      $\Large\displaystyle\begin{gathered} A_{2} = x + r = -\frac{1}{2} + \frac{1}{3} = -\frac{1}{6}\end{gathered}$

      $\Large\displaystyle\begin{gathered} A_{3} = x + 2r = -\frac{1}{2} + 2\cdot\frac{1}{3} = \frac{1}{6}\end{gathered}$

✅ Portanto, a P.A. é:

                     $\Large\displaystyle\begin{gathered} P.A.\bigg(-\frac{1}{2},\,-\frac{1}{6},\,\frac{1}{6}\bigg)\end{gathered}$

 

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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