Question

em uma pa, sabe se que a a1 + a3= -6 e a2 + a6=2. Seu 10 termo é igual a

Answer 1

Bom dia!

Solução!

$a1+a3=-6\\ a2+a6=2\\\\\\\ a1+a1+2r=-6\\ a1+r+a1+5r=2\\\\\\\ 2a1+2r=-6\\ 2a1+6r=2\\\\\\\\\\ 2a1+2r=-6.(-1)\\ 2a1+6r=2\\\\\\\\\\ -2a1-2r=6\\ ~~~~2a1+6r=2\\\\\\\\\\ -2r+6r=6+2\\\\\\\ 4r=8\\\\ r= \dfrac{8}{4}\\\\ \boxed{r=2}$


$2a1+6r=2\\\\\ 2a1+6(2)=2\\\\\ 2a1+12=2\\\\\\ 2a1=2-12\\\\\ 2a1=-10\\\\\ a1= \dfrac{-10}{2} \\\\ \boxed{a1=-5}$



$a10=a1+9r\\\\\ a10=-5+9(2)\\\\\ a10=-5+18\\\\\ \boxed{a10=13}$



$\boxed{Resposta: Decimo~~termo~~e\´~~13}$


Bom dia!
Bons estudos!

Answer 2

Resposta:

Resposta: D) 13

Explicação passo a passo:

a1 + a3 = -6

a1 + a1 +2r = -6

2a1 + 2r = -6 (I)

a2 + a6 = 2

a1 + r + a1 +5r = 2

2a1 + 6r = 2 (II)

(sistema)

2a1 + 2r = -6   .(-1)  

2a1 + 6r = 2  

(sistema)

-2a1 - 2r = 6

2a1 + 6r = 2

         4r = 8

           r = 8/4

           r = 2

2a1 + 2r = - 6

2a1 + 2.2 = - 6

2a1 + 4 = - 6

2a1 = - 6 – 4

a1 = -10/2

a1 = -5

Assim, r = 2 e a1 = -5

a10 = a1 + 9r

a10 = -5 + 9.2

a10 = -5 + 18

a10 = 13