Respostas

  • Usuário do Brainly
2013-06-26T01:42:03-03:00

a) Observe que:

 

\text{q}=\dfrac{\text{a}_2}{\text{a}_1}=\dfrac{\text{a}_3}{\text{a}_2}=\dots=\dfrac{\text{a}_{\text{n}}}{\text{a}_{\text{n}-1}}

 

Desta maneira, temos:

 

\dfrac{\text{x}+2}{\text{x}}=\dfrac{2\text{x}+4}{\text{x}+2}

 

(\text{x}+2)\cdot(x+2)=\text{x}\cdot(2\text{x}+4)

 

\text{x}^2+4\text{x}+4=2\text{x}^2+4\text{x}

 

Donde, segue:

 

\text{x}^2=4

 

\sqrt{\text{x}}=\sqrt{4}

 

\text{x}=\pm2

 

\text{S}=\{-2, 2\}

 

Observe que, \text{x}\ne-2, porque, teríamos:

 

-2, 0, -4 absurdo -_-

 

Logo, \text{x}=2.

 

 

b) Observe que:

 

\text{a}_{\text{n}}=\text{a}_1\cdot\text{q}^{\text{n}-1}

 

Segundo o enunciado, \text{a}_7=32 e \text{q}=2.

 

Desta maneira:

 

\text{a}_7=\text{a}_1\cdot\text{q}^{7-1}

 

32=\text{a}_1\cdot2^6

 

Logo:

 

\text{a}_1=\dfrac{32}{2^6}=\dfrac{2^5}{2^6}=2^{5-6}=2^{-1}=\dfrac{1}{2}

 

Portanto, o primeiro termo desta P.G. é \dfrac{1}{2}