Question

interpole cincos meios geométricos entre 4 e 2916

Answer 1

a1 = 4
an = 2916
n = 5+2
n = 7

an = a1.q^(n-1)
2916 = 4.q^(7-1)
2916 = 4.q^6
2916/4 = q^6
q^6 = 729
q^6 = 3^6
q = 3

PG = (4, 12, 36, 108, 324, 972, 2916)

0729|3
0243|3
0081|3
0027|3
0009|3
0003|3
0001 / 2916 = 2²x3^6 = 4 x 3^6