Question

Inserir 3 meios geométricos entre 6 e 1536?

Answer 1

Ola Matheus

PG

u1 = 6
u5 = u1*q^4 = 1536

6*q^4 = 1536
q^4 = 1536/6 = 256
q = 4

u1 = 6
u2 = u1q = 6*4 = 24
u3 = u2q = 24*4 = 96
u4 = u3q = 96*4 = 384 
u5 = u4q = 384*4 = 1536

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