Question

Insira cinco meio geométricos ente 10 e 20 nessa ordem.

Answer 1

Olá Llopesam,


Para inserir 5 meios geométricos entre 10 e 20, iremos considerar o 10 o primeiro termo e o 20 o sétimo termo

Equação geral de uma P.G é $a_n=a_1\cdot q^{n-1}$

Encontrando a razão:

$20=10\cdot q^{7-1}\\20=10\cdot q^6\\\\\frac{20}{10}=q^6\\\\10=q^6\\\\q=\sqrt[6]{10}$

Encontrando os termos:

$a_2=10\cdot (\sqrt[6]{10})^{2-1}\\a_2=10\cdot \sqrt[6]{10}\\a_2=10\cdot 10^{\frac{1}{6}}\\a_2=10^{\frac{7}{6}}\\a_2=\sqrt[6]{10^7}\\a_2=\sqrt[6]{10^{6+1}}\\a_2=10\sqrt[6]{10}\\\\a_3=10\cdot (\sqrt[6]{10})^{3-1}\\a_3=10\cdot 10^{\frac{2}{6}}\\a_3=10^{\frac{8}{6}}\\a_3=\sqrt[4]{10^3}\\\\a_4=10\cdot (\sqrt[6]{10})^{4-1}\\a_4=10\cdot 10^{\frac{3}{6}}\\a_4=10^{\frac{3}{2}}\\a_4=\sqrt{10^{2+1}}\\a_4=\sqrt{10^2.10}\\a_4=10\sqrt{10}$

$a_5=10\cdot (\sqrt[6]{10})^{5-1}\\a_5=10\cdot 10^{\frac{4}{6}}\\a_5=10^\frac{5}{3}\\a_5=\sqrt[3]{10^5}\\a_5=\sqrt[3]{10^{3+2}}\\a_5=10\sqrt[3]{10^2}$

$a_6=10\cdot (\sqrt[6]{10})^{6-1}\\a_6=10\cdot 10^\frac{5}{6}\\a_6=10^{\frac{11}{6}}\\a_6=\sqrt[6]{10^{11}}\\a_6=\sqrt[6]{10^{6+5}}\\a_6=10\sqrt[6]{10^5}$

$\\\\\mathsf{P.G=(10,~10\sqrt[6]{10},~\sqrt[4]{10^3},~10\sqrt{10},~10\sqrt[3]{10^2},~10\sqrt[6]{10^5},~20})$

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