Question

Considere a p.a. em que

a6+a8=40
a2+a4=8

a) razão da p.a.
B) formar a p.a. De 9 termos

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\begin{cases}\sf{a_6 + a_8 = 40}\\\sf{a_2 + a_4 = 8}\end{cases}$

$\begin{cases}\sf{a_1 + 5r + a_1 + 7r = 40}\\\sf{a_1 + r + a_1 + 3r = 8}\end{cases}$

$\begin{cases}\sf{2a_1 + 12r = 40}\\\sf{2a_1 + 4r = 8\:\:\times (-1)}\end{cases}$

$\begin{cases}\sf{2a_1 + 12r = 40}\\\sf{-2a_1 - 4r = -8}\end{cases}$

$\sf{8r = 32}$

$\boxed{\boxed{\sf{r = 4}}}$

$\sf{2a_1 + 4r = 8}$

$\sf{2a_1 + 16 = 8}$

$\sf{2a_1 = -8}$

$\boxed{\boxed{\sf{a_1 = -4}}}$

$\boxed{\boxed{\sf{S = \{-4,0,4,8,12,16,20,24,28\}}}}$