Question

ache a soma dos 40 primeiros termos da PA (8,2, ...)

Answer 1

Razão r = 2-8 = -6

$a_{n} = a_{1} + (n-1)r \\ \\ a_{40} = 8 + (40-1)(-6) \\ \\ a_{40} = 8 + 39(-6) \\ \\ a_{40} = 8 + (-234) \\ \\ a_{40} = 8 -234 \\ \\ a_{40} = -226$

$S_{n} = \frac{(a_{1}+a_{n})n}{2} \\ \\ S_{40} = \frac{(8-226)40}{2} \\ \\ S_{n} = -218 . 20 \\ \\ S_{n} = -4360$

Answer 2

r= 2 - 8 = - 6

an= a1 + (n - 1) • r

a40= 8 + (40 - 1) • (-6)

a40= 8 + 39 • (-6)

a40= 8 - 234

a40= - 226

sn= (a1 + an) • n/2

s40= (8 + ( - 226)) • 40/2

s40= - 218 • 40/2

s40= - 109 • 40

s40= - 4360