PA de 8 termos em que a1=6 e r= -4
Soluções para a tarefa
Respondido por
5
a1 = 6
r = -4
n an = a1 + ( n -1) * r an
1 an = 6 + ( 1 -1) * -4 6
2 an = 6 + ( 2 -1) * -4 2
3 an = 6 + ( 3 -1) * -4 -2
4 an = 6 + ( 4 -1) *-4 -6
5 an = 6 + ( 5 -1) *-4 -10
6 an = 6 + ( 6 -1) *-4 -14
7 an = 6 + ( 7 -1) *-4 -18
8 an = 6 + ( 8 -1) * -4 -22
PA = {6, 2, -2, -6, -10, -14, -18, -22}
r = -4
n an = a1 + ( n -1) * r an
1 an = 6 + ( 1 -1) * -4 6
2 an = 6 + ( 2 -1) * -4 2
3 an = 6 + ( 3 -1) * -4 -2
4 an = 6 + ( 4 -1) *-4 -6
5 an = 6 + ( 5 -1) *-4 -10
6 an = 6 + ( 6 -1) *-4 -14
7 an = 6 + ( 7 -1) *-4 -18
8 an = 6 + ( 8 -1) * -4 -22
PA = {6, 2, -2, -6, -10, -14, -18, -22}
Helvio:
De nada.
Respondido por
2
PA=a₁+(1+n)r
PA=6+(1+8)-4
PA=6+9×-4
PA=6-36
PA=-30
PA=6+(1+8)-4
PA=6+9×-4
PA=6-36
PA=-30
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