Question

interpole oito meios geometricos entre 2000 e 125 / 32

Answer 1

a1 = 2 000
a10 = 125/32
q = ?

a10 = a1 * q^9
125/32 = 2 000 * q^9
q^9 = 125/32 / 2000
q^9 = 125/32 * 1/2000
q^9 = 125/64000
q = 1/2


P.G. (2000; 1000; 500; 250; 125; 125/2; 125/4; 125/8; 125/16; 125/32)