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se considera progresia geometrica in care b2=8 si b4=32.sa se calculeze b6

Răspuns :

[tex]\underline{b_n=b_1*q^{n-1}} \\\\\\ b_6=b_4*q^2 \\\\\\ b_2=b_1*q \\ b_4=b_1*q^3 \\\\ q^2= \frac{\not b1*q^3}{\not b_1*q}=\frac{b_4}{b_2} = \frac{32}{8} \to 4 \\\\ \underline{q^2=4} \\\\\\ b_6=32*4 \\\\ \boxed{b_6=128}[/tex]