Question

75- Dada a PG (2/5.4/15.8/45..), identifique a posição do termo 256/10,935.

Mim ajudem pelo amor de Deus..

Answer 1

Fórmulas de P.G:

$a_{n}=a_{1}*q^{n-1}$

q = Razão da P.G

$q = a_{2}/a_{1}=a_{3}/a_{2}=a_{4}/a_{3}=...=a_{n}/a_{n-1}$

Lembre-se:

$a^{n}/b^{n}=(a/b)^{n}\\a^{x}=a^{y}<=>x=y$

___________________________

$P.G~(2/5,~4/15,~8/45,...)$

$a_{1}=2/5\\a_{2}=4/15\\\\q=a_{2}/a_{1}=(4/15)/(2/5)=(4/15)*(5/2)=2/3$

$a_{n}=a_{1}*q^{n-1}\\a_{n}=(2/5)*(2/3)^{n-1}\\256/10935=(2/5)*(2/3)^{n-1}\\(256/10935)/(2/5)=(2/3)^{n-1}\\(256/10935)*(5/2)=(2/3)^{n-1}\\128/2187=(2/3)^{n-1}$

Veja que 128 = 2⁷, e 2187 = 3⁷. Substituindo:

$2^{7}/3^{7}=(2/3)^{n-1}\\(2/3)^{7}=(2/3)^{n-1}$

Bases iguais, iguale os expoentes:

$7 = n - 1\\7+1=n\\n=8$

R: 256/10935 é o oitavo termo da P.G