Question

Em uma P.G., a5=24 e a8=3. Obtenha:

A) A razão da P.G.
B) O décimo termo da P.G.
C) A soma dos 8 primeiro termos

Answer 1

Explicação passo-a-passo:

a)

Utilizando a fórmula do termo geral:

$\sf a_n=a_1\cdot q^{n-1}$

=> a5 = 24

$\sf a_5=a_1\cdot q^4$

$\sf a_1\cdot q^4=24$

=> a8 = 3

$\sf a_8=a_1\cdot q^7$

$\sf a_1\cdot q^7=3$

$\sf a_1\cdot q^4\cdot q^3=3$

Substituindo $\sf a_1\cdot q^4~por~24$:

$\sf 24\cdot q^3=3$

$\sf q^3=\dfrac{3}{24}$

$\sf q^3=\dfrac{1}{8}$

$\sf q=\sqrt[3]{\dfrac{1}{8}}$

$\sf \red{q=\dfrac{1}{2}}$

Substituindo em $\sf a_1\cdot q^4=24$:

$\sf a_1\cdot\Big(\dfrac{1}{2}\Big)^4=24$

$\sf a_1\cdot\dfrac{1}{16}=24$

$\sf a_1=16\cdot24$

$\sf \red{a_1=384}$

b)

Utilizando a fórmula do termo geral:

$\sf a_n=a_1\cdot q^{n-1}$

$\sf a_{10}=a_1\cdot q^9$

$\sf a_{10}=384\cdot\Big(\dfrac{1}{2}\Big)^9$

$\sf a_{10}=384\cdot\dfrac{1}{512}$

$\sf a_{10}=\dfrac{384\div128}{512\div128}$

$\sf \red{a_{10}=\dfrac{3}{4}}$

c)

A soma dos n primeiros termos de uma PG é dada por:

$\sf S_n=\dfrac{a_1\cdot(q^n-1)}{q-1}$

$\sf S_8=\dfrac{384\cdot\Big[\Big(\frac{1}{2}\Big)^8-1\Big]}{\frac{1}{2}-1}$

$\sf S_8=\dfrac{384\cdot\Big[\frac{1}{256}-1\Big]}{\frac{1-2}{2}}$

$\sf S_8=\dfrac{384\cdot\Big[\frac{1-256}{256}\Big]}{-\frac{1}{2}}$

$\sf S_8=\dfrac{384\cdot\Big[-\frac{255}{256}\Big]}{-\frac{1}{2}}$

$\sf S_8=\dfrac{-\frac{97920}{256}}{-\frac{1}{2}}$

$\sf S_8=-\dfrac{97920}{256}\cdot\Big(-\dfrac{2}{1}\Big)$

$\sf S_8=\dfrac{195840}{256}$

$\sf \red{S_8=765}$