Question

5 meios geometricos entri 1 e 64​

Answer 1

Se inserirmos 5 termos entre 1 e 64, teremos uma PG de 7 termos.

{1 , a₂ , a₃ , a₄ , a₅ , a₆ , 64}

Vamos então determinar a razão (q) desta PG com auxilio da equação do termo geral da PG:

$a_n~=~a_1~.~q^{n-1}\\\\\\64~=~1~.~q^{7-1}\\\\\\q^{6}~=~64\\\\\\q~=~\sqrt[6]{64}\\\\\\\boxed{q~=~2}$

Podemos agora determinar o valor dos termos inseridos.

$a_2~=~a_1~.~q~=~1~.~2~~~\rightarrow~~~\boxed{a_2~=~2}\\\\\\a_3~=~a_2~.~q~=~2~.~2~~~\rightarrow~~~\boxed{a_3~=~4}\\\\\\a_4~=~a_3~.~q~=~4~.~2~~~\rightarrow~~~\boxed{a_4~=~8}\\\\\\a_5~=~a_4~.~q~=~8~.~2~~~\rightarrow~~~\boxed{a_5~=~16}\\\\\\a_6~=~a_5~.~q~=~16~.~2~~~\rightarrow~~~\boxed{a_6~=~32}\\\\\\\\\boxed{PG:~~ \{1~,~2~,~4~,~8~,~16~,~32~,~64\}}$