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2014-08-18T23:20:41-03:00
Olá Letícia,

\begin{cases}a_2=12\\
a_1+a_{3}=-51\end{cases}
\begin{cases}a_1\cdot q=12\\
a_1+(a_1\cdot q^2)=-51 \end{cases}\begin{cases}a_1= \dfrac{12}{q}~~(I)\\
a_1+(a_1\cdot q^2)=-51~~(II) \end{cases}\\\\
Substitua~I~em~II:\\\\
 \dfrac{12}{q}+\left( \dfrac{12}{q}\cdot q^2\right)=-51\\\\\\
 \dfrac{12}{q}+ \dfrac{12q^2}{q}=-51\\\\\\
 \dfrac{12}{q}+12q=-51\\\\\\
12+12q^2=-51q\\
12q^2+51q+12=0

\Delta=51^2-4\cdot12\cdot12\\
\Delta=2.601-576\\
\Delta=2.025\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ q'=\dfrac{-51+45}{24}= \dfrac{-6}{24}= -\dfrac{1}{4}   \\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nearrow\\
q= \dfrac{-51\pm \sqrt{2.025} }{2\cdot12}= \dfrac{-51\pm45}{24}\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\searrow~\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~q''= \dfrac{-51-45}{24}= \dfrac{-96}{~~24}=-4

Portanto esta P.G. possui duas razões,  - \dfrac{1}{4}~~e~~-4 .

Tenha ótimos estudos ;D


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