Question

calcular a soma dos termos da P.A ( -16 ,-14 ,-12,...,64)

Answer 1

1° vamos descobrir quantos termos tem essa PA:
$P.A ( -16 ,-14 ,-12,...,64)\\\\\\\\ a_1=-16\\ an=64\\ r=2\\ n=?\\\\\\ a_n=a_1+(n-1)*r\\\\\\ 64=-16+(n-1)*2\to \\\\ 64=-16+2n-2\to \\\\ 64+16+2=2n\to \\\\82=2n\to \\\\ \frac{82}{2} =n\to \\\\ n=41$


Agora passamos à soma de termos:

$S_n= \frac{(a_1+a_n)*n}{2} \\\\ S_{41}= \frac{(-16+64)*41}{2} \to \\\\ S_{41}= \frac{48*41}{2} \to \\\\ S_{41}= \frac{1968}{2} \to \\\\ S_{41}=984$


R.: A soma dos termos dessa PA é igual 936.


Espero ter ajudado...

Answer 2

r = a2 - a1 => -14 - (-16) = -14 + 16 = 2

an = a1 + (n -1)*r

64 = -16 + (n -1)* 2
64+16 =(n -1)*2
80/2 = n -1
40 +1 = n
41 = n

$sn= \frac{(a1 + an) *n}{2} = \frac{(-16+64)*41}{2}= \frac{48*41}{2}= \frac{1968}{2} = 984$