Question

determine o 7 termo da pg 27,9,3

Answer 1

Dada a P.G:

$\mathsf{\left(27,\:9,\:3,...\right)}$

Calculando a razão:

$\\\\\\\mathsf{q=\dfrac{a_n}{a_{n-1}}}\\\\\\\mathsf{q=\dfrac{9}{27}}\\\\\\\mathsf{q=\dfrac{1}{3}}$

Substituindo na formula geral de uma P.G:

$\mathsf{a_n=a_k\cdot q^{n-k}}\\\\\\\mathsf{a_7=a_1\cdot q^{\left(7-1\right)}}\\\\\\\mathsf{a_7=27\cdot \left(\dfrac{1}{3}\right)^6}\\\\\\\mathsf{a_7=\dfrac{27}{3^6}}\\\\\\\mathsf{a_7=\dfrac{3^3}{3^6}}\\\\\\\mathsf{a_7=\dfrac{1}{3^3}}\\\\\\\boxed{\mathsf{a_7=\frac{1}{27}}}\: \: \checkmark$

Answer 2

q = a2/a1
q = 9/27
q = 1/3

an = a1 . q^n-1
a7 = a1 . q^7-1
a7 = 27 . (1/3)^6
a7 = 27 . 1/729
a7 = 27/729
a7 = 1/27