Question

insira quatro meios geométricos entre 1 e 3125

Answer 1

a1 = 1

an = 3125

N = 4 + 2  ⇒  N = 6

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an = a1 . q ^n-1

3125 = 1. q ^5

5^5  = q ^5

q = 5

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PG ( 1 , 5 , 25 , 125 , 625 , 3125 )

Answer 2

P.G.(1,5,25,125,625,3125)
a 1 = 1
a 6 = a 1 . q^n- 1
3125 = 1 . q^5
q^5 = 3125
q^5 = 5^5
q = 5
a 2 = a 1 . 5¹
a 2 = 1.5 = 5
a 3 = 5 . 5 = 25
a 4 = 25 . 5 = 125
a 5 = 125 . 5 = 625
a 6 = 625 . 5 = 3125