Question

Pfvvv Alguém me ajuda: Insira seis meios geométricos entre 2/3 e 1458​
​

Answer 1

Inserindo-se outros 6 termos, teremos uma PG com 8 termos.

{ 2/3 , a2 , a3 , a4 , a5 , a6 , a7 , 1458 }

Vamos, então, começar determinando a razão da PG utilizando a equação do termo geral:

$\boxed{a_n~=~a_1~.~q^{n-1}}\\\\\\a_8~=~a_1~.~q^{8-1}\\\\\\1458~=~\frac{2}{3}~.~q^{7}\\\\\\q^{7}~=~1458~.~\frac{3}{2}\\\\\\q^{7}~=~2187\\\\\\q~=~\sqrt[7]{2187}\\\\\\\boxed{q~=~3}$

Agora podemos determinar quanto vale cada meio geométrico adicionado, acompanhe:

$a_2~~=~~a_1~.~q~~=~~\frac{2}{3}~.~3~~~\rightarrow~~\boxed{a_2~=~2}\\\\\\a_3~~=~~a_2~.~q~~=~~2~.~3~~~\rightarrow~~\boxed{a_3~=~6}\\\\\\a_4~~=~~a_3~.~q~~=~~6~.~3~~~\rightarrow~~\boxed{a_4~=~18}\\\\\\a_5~~=~~a_4~.~q~~=~~18~.~3~~~\rightarrow~~\boxed{a_5~=~54}\\\\\\a_6~~=~~a_5~.~q~~=~~54~.~3~~~\rightarrow~~\boxed{a_6~=~162}\\\\\\a_7~~=~~a_6~.~q~~=~~162~.~3~~~\rightarrow~~\boxed{a_7~=~486}$

Answer 2

resolucao!

an = a1 * q^n - 1

1458 = 2/3 * q^7

1458 / 2/3 = q^7

1458 * 3/2 = q^7

4374/2 = q^7

2187 = q^7

3^7 = q^7

q = 3

PG = { 2/3 , 2 , 6 , 18 , 54 , 162 , 486 , 1458 }