Question

interpole 4 meios geometricos entre 4 e 128

Answer 1

[/tex]a n= a_{1} . q^{n-1} 128 = 4. q^{6-1} 128 = 4 . q_{5} 128: 4 = q_{5}
32= q_{5} \sqrt[5]{32} = q q = 2 (4,8,16,32,64, 128)[/tex]