Question

determine a soma dos 12 primeiros termos da p.a
a) (8,11,14,17...)
b) (0,12,24)

Answer 1

a)
an = a1 + (n-1) . r
an = 8 + (12-1) . 3
an = 8 + 11.3
an = 8+33
an = 41

$sn = \frac{(a1 + an) \times n}{2}$

$sn = \frac{(8 + 41) \times 12}{2}$

simplifica o 12 com o 2 (dividindo eles) = 6
sn = 49*6
sn = 294

b)
an = 0 + (12-1) . 12
an = 11*12
an = 132

$sn = \frac{(a1 + an) \times n}{2}$

$sn = \frac{(0 + 132) \times 12}{2}$

sn = 132*6
sn = 792