Question

calcule a soma dos doze primeiros termos da p.a. (3/4, 39/20, 63/20, ...)

Answer 1

Sendo a1 = 3/4 = 15/20, a2 = 39/20, temos:

an = a1 + (n-1).r
a2 = a1 + r
39/20 = 15/20 + r
r = 24/20 = 12/10

Assim, temos:

a12 = a1 + 11r
a12 = 15/20 + 264/20 = 279/20

Então, a soma dos 12 primeiros termos será:

S p.a. = (a1 + a12).n/2
S p.a. = (15/20 + 279/20). 12/2
S p.a. = 1764/20 = 88,2

Answer 2

a1 = 3/4
r = a2-a1
r = 39/20 - 3/4
r = 39/20 - 15/20
r = 24/20 = 12/10 = 6/5
n = 12
a12 = a1+11.r
a12 = 3/4 + 11.6/5
a12 = 3/4 + 66/5
a12 = 15/20 + 264/20
a12 = 279/20
Sn = (a1+an).n/2
Sn = (3/4+279/20).12/2
Sn = (15/20+279/20).6
Sn = (294/20).6
Sn = 147/10 . 6
Sn = 882/10
Sn = 441/5