Jeżeli ciąg (an) jest ciągiem geometrycznym o ilorazie q, to n-ty wyraz ciągu określony jest wzorem: $a_n=a_1\cdot q^{n-1}$

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a)

$a_1=-2,q=\sqrt{3}$

Wobec tego:

$a_2=a_1\cdot q^{2-1}=-2\cdot {\left(\sqrt{3}\right)}^1=-2\sqrt{3}$

$a_3=a_1\cdot q^{3-1}=-2\cdot {\left(\sqrt{3}\right)}^2=-2\cdot 3=-6$    

$a_4=a_1\cdot q^{4-1}=-2\cdot {\left(\sqrt{3}\right)}^3=-2\cdot 3\sqrt{3}=-6\sqrt{3}$  

$a_5=a_1\cdot q^{5-1}=-2\cdot {\left(\sqrt{3}\right)}^4=-2\cdot 9=-18$  

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b)

$a_1=\sqrt{2},q=1-\sqrt{2}$

Wobec tego:

$a_2=a_1\cdot q^{2-1}=\sqrt{2}\cdot {\left(1-\sqrt{2}\right)}^1=\sqrt{2}-2$

$a_3=a_1\cdot q^{3-1}=\sqrt{2}\cdot {\left(1-\sqrt{2}\right)}^2=\sqrt{2}\left(1-2\sqrt{2}+2\right)=\sqrt{2}\left(3-2\sqrt{2}\right)=3\sqrt{2}-4$    

$a_4=a_1\cdot q^{4-1}=\sqrt{2}\cdot {\left(1-\sqrt{2}\right)}^3=\sqrt{2}\left(1-3\sqrt{2}+6-2\sqrt{2}\right)=\sqrt{2}\left(7-5\sqrt{2}\right)=7\sqrt{2}-10$  

$a_5=a_1\cdot q^{5-1}=\sqrt{2}\cdot {\left(1-\sqrt{2}\right)}^4=\sqrt{2}{\left(1-\sqrt{2}\right)}^2{\left(1-\sqrt{2}\right)}^2=$

$=\sqrt{2}\left(1-2\sqrt{2}+2\right)\left(1-2\sqrt{2}+2\right)=\sqrt{2}{\left(3-2\sqrt{2}\right)}^2=\sqrt{2}\left(9-12\sqrt{2}+8\right)=$   

$=\sqrt{2}\left(17-12\sqrt{2}\right)=17\sqrt{2}-24$  

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c)

$a_1=\frac{1}{3},q=-1$

Wobec tego:

$a_2=a_1\cdot q^{2-1}=\frac{1}{3}\cdot {\left(-1\right)}^1=-\frac{1}{3}$

$a_3=a_1\cdot q^{3-1}=\frac{1}{3}\cdot {\left(-1\right)}^2=\frac{1}{3}$    

$a_4=a_1\cdot q^{4-1}=\frac{1}{3}\cdot {\left(-1\right)}^3=-\frac{1}{3}$  

$a_5=a_1\cdot q^{5-1}=\frac{1}{3}\cdot {\left(-1\right)}^4=\frac{1}{3}$  

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d)

$a_1=3,a_2=-6$

Wobec tego:

$q=\frac{a_2}{a_1}=\frac{-6}{3}=-2$

$a_3=a_1\cdot q^{3-1}=3\cdot {\left(-2\right)}^2=3\cdot 4=12$    

$a_4=a_1\cdot q^{4-1}=3\cdot {\left(-2\right)}^3=3\cdot \left(-8\right)=-24$  

$a_5=a_1\cdot q^{5-1}=3\cdot {\left(-2\right)}^4=3\cdot 16=48$