Question

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

Answer 1

$a_{n} = a_{1} + (n - 1).r; a_{40} = 8 + (40 - 1). (-6) = 8 + 39. (-6)= - 226;$; $S_{n} = ( a_{1} + a_{n} ).n/2 = (8 - 226). \frac{40}{2} = -218 . 20 --\ \textgreater \ S_{40} = - 4.360$

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