Question

Calcule a soma dos doze primeiros termos da P. A (-3, -1, 1, 3, ...).
a- 93.
b- 94.
c- 95.
d- 96.
e- 97.

Answer 1

$(a_1,a_2,a_3,a_4,\dots)=(-3,-1,1,3,\dots)$

$\boxed{a_n=a_k+(n-k)r}$

$a_2=a_1+r\ \therefore\ r=a_2-a_1=-1-(-3)\ \therefore\ \boxed{r=2}$

$a_{12}=a_1+11r=-3+11(2)\ \therefore\ \boxed{a_{12}=19}$

$\boxed{S_{n}=\dfrac{n}{2}\big(a_1+a_n\big)}$

$S_{12}=\dfrac{12}{2}\big(a_1+a_{12}\big)=6\big(-3+19\big)\ \therefore\ \boxed{S_{12}=96}$

Letra D.

Answer 2

Resposta:

D

Explicação passo-a-passo:

an = a1 + (n – 1)r

a12 = -3 + (12 – 1).2

a12 = -3 + (11).2

a12 = -3+22

a12 = 19

Sn = n(a1 + an)

      2

S12= 12(-3+ 19)

        2

S12 = 12(16)

         2

S12 = 192

         2

S12 = 96