Będziemy korzystać ze wzoru na n-ty wyraz ciągu geometrycznego o ilorazie q.

$a_n=a_1\cdot q^{n-1}$

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$a)a_2=a_1\cdot q=27\cdot \frac{2}{3}=9\cdot 2=18$  

$a_3=a_2\cdot q=18\cdot \frac{2}{3}=6\cdot 2=12$  

$a_4=a_3\cdot q=12\cdot \frac{2}{3}=4\cdot 2=8$  

$a_6=a_4\cdot q^2=8\cdot {\left(\frac{2}{3}\right)}^2=8\cdot \frac{4}{9}=\frac{32}{9}=3\frac{5}{9}$  

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$b)a_2=a_1\cdot q=9\cdot \left(-1\right)=-9$  

$a_3=a_2\cdot q=-9\cdot \left(-1\right)=9$  

$a_4=a_3\cdot q=9\cdot \left(-1\right)=-9$  

$a_6=a_4\cdot q^2=-9\cdot {\left(-1\right)}^2=-9$  

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

$a_3=a_2\cdot q=\left(2-\sqrt{2}\right)\cdot \sqrt{2}=2\sqrt{2}-2$  

$a_4=a_3\cdot q=\left(2\sqrt{2}-2\right)\cdot \sqrt{2}=2\cdot 2-2\sqrt{2}=4-2\sqrt{2}$  

$a_6=a_4\cdot q^2=\left(4-2\sqrt{2}\right)\cdot {\left(\sqrt{2}\right)}^2=\left(4-2\sqrt{2}\right)\cdot 2=8-4\sqrt{2}$