Question

ache a soma dos 10 primeiros termos da P.A (8,2, ...)

Answer 1

r= a2-a1
r= 2-8
r= -6

an= a1+(n-1).r
a10= 8 +(10-1).-6
a10= -46

$Sn= \frac{(a1+an).n}{2}$
$S10= \frac{(8-46).10}{2}$
$S10= \frac{-38*10}{2}$
$S10= \frac{-380}{2}$
$\boxed{\boxed{S10= -190}}$

Answer 2

$a_{10} = 8 + (10 - 1). (-6)=8 + 9. (-6)=8 - 54--\ \textgreater \ a_{10} = -46;$ $S_{10} = \frac{(8-46).10}{2} = - 38.5--\ \textgreater \ S_{10} = -190.$