Resposta :

 

2 = a1.(1/4)^(6-1)

 

2 = a1(1/4)^5

 

2 = a1(2^-2)^5

 

2 = a1.2^-10

 

a1 =    2     =>  a1 = 2 . 2^10 => a1 = 2^11

          2^-10 

Observe que:

 

[tex]\text{a}_{\text{n}}=\text{a}_1\cdot\text{q}^{\text{n}-1}[/tex]

 

Onde, [tex]\text{q}=\dfrac{\text{a}_2}{\text{a}_1}=\dfrac{\text{a}_3}{\text{a}_2}=\dots=\dfrac{\text{n}}{\text{a}_{\text{n}-1}}[/tex]

 

Segundo o enunciado, [tex]\text{a}_{\text{n}}=2[/tex], [tex]\text{n}=6[/tex], [tex]\text{q}=\dfrac{1}{4}[/tex], logo:

 

[tex]2=\text{a}_1\cdot\left(\dfrac{1}{4}\right)^{6-1}[/tex]

 

Observe que, [tex]\dfrac{1}{4}=2^{-2}[/tex], desta maneira:

 

[tex]2=\text{a}_1\cdot(2^{-2})^{6-1}[/tex]

 

[tex]2=\text{a}_1\cdot(2^{-2})^5[/tex]

 

[tex]2=\text{a}_1\cdot2^{-10}[/tex]

 

Donde, obtemos:

 

[tex]\text{a}_1=\dfrac{2}{2^{-10}}=2^{1-(-10)}=2^{1+10}=2^{11}[/tex]

 

Logo, o primeiro termo desta P.G. é [tex]2^{11}=2~048[/tex].

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