Question

determine a soma dos termos da PA finita(6,10,14,...,134).

Answer 1

$an=a1+(n-1).r \\ \\ 134=6+(n-1).4 \\ \\ 134-6=(n-1).4 \\ \\ 128=(n-1).4 \\ \\ \frac{128}{4}=n-1 \\ \\ 32=n-1 \\ \\ n=33$



$Sn= \frac{(a1+an).n}{2} \\ \\ Sn= \frac{(6+134).31}{2} \\ \\ Sn= \frac{(140).31}{2} \\ \\ Sn=70.31 \\ \\ Sn=2170$

Answer 2

6,10,14,18,22,26,30,34,38,42,46,50,54,58,62,66,70,74,78,82,86,90,94,98,102,106,110,114,118,122,126,130,134, 
 espero ter ajudado :)