Question

2) Calcule a razão da PG (2,__,___,___,___, 192).





3)DESAFIO : Seja a PG (____, 4, ____, 64,____) Calcule a razão e os termos que faltam.
pfvr me ajudem pfvr​

Answer 1

Resposta:

2) A razão é, aproximadamente, 2,49146

3) A P.G. é igual a (1, 4, 16, 64, 256)

Explicação passo-a-passo:

A equação para o termo geral de uma P.G. é dada por

$a_n=a_1\;.\;q^{(n-1)}$

2) Calcule a razão da P.G. (2,__,___,___,___, 192).

$n=6\\a_1=2\\a_6=192\\\\\\a_n=a_1\;.\;q^{(n-1)}\\\\a_6=a_1\;.\;q^{(6-1)}\\\\192=2\;.\;q^{(6-1)}\\\\192=2\;.\;q^5\\\\q^5=\dfrac{192}{2}\\\\q^5=96\\\\q=\sqrt[5]{96}\\\\\boxed{q\approx2{,}49146}$

3) Seja a PG (____, 4, ____, 64,____) Calcule a razão e os termos que faltam.

$a_1=\;?\\a_2=4\\a_4=64\\\\\\a_2=a_1\;.\;q^{(2-1)}\\\\4=a_1\;.\;q^1\\\\4=a_1\;.\;q\\\\a_1=\dfrac{4}{q}\\\\\\a_4=a_1\;.\;q^{(4-1)}\\\\64=a_1\;.\;q^3\\\\64=\dfrac{4}{q}\;.\;q^3\\\\64=4\;.\;q^2\\\\q^2=\dfrac{64}{4}\\\\q^2=16\\\\q=\sqrt{16}\\\\q=4$

Portanto,

$a_1=\dfrac{4}{q}=\dfrac{4}{4}=1\\\\a_3=1\;.\;4^{(3-1)}=1\;.\;4^2=1\;.\;16=16\\\\a_5=1\;.\;4^{(5-1)}=1\;.\;4^4=1\;.\;256=256$

Logo, a P.G. é igual a (1, 4, 16, 64, 256)