Question

na PA ( 68, 62, 56,50,...), a soma de seus.
 
 A) seis primeiros termos 
 B) quatro ultimo termo, admitindo que a sequencia tem 10 termos.

Answer 1

a1 = 68
r = -6
$S(n)= \frac{(a1+an)n}{2}$
$a(n)=a1+ (n-1)r$

A)
$a(6)=68+5.(-6)=\ \textgreater \ 38$

$S(6)= \frac{(68+38).6}{2} =\ \textgreater \ {106.3} =\ \textgreater \ 318$

B)
$a(10)=68+9(-6)=\ \textgreater \ 14$

$S(10)= \frac{(68+14)10}{2}=\ \textgreater \ 82.5=\ \textgreater \ 410$