Question

interpole oito meios geométricos entre 2000 e
$\frac{125}{32}$
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Answer 1

resolução!

an = a1 * q^n - 1

125/32 = 2000 * q^9

125/32 ÷ 2000 = q^9

125/32 * 1/2000 = q^9

125/64000 = q^9

1/512 = q^9

(1/2)^9 = q^9

q = 1/2

PG = { 2000 , 1000 , 500 , 250 , 125 , 125/2 , 125/4 , 125/8 , 125/16 , 125/32 }