Question

numa PA a2+a4=16 e a5+a7=28. determine a soma dos 28 primeiros termos dessa PA

Answer 1

a₂ + a₄ = 16

a₅ + a₇ = 28

a₁ + r + a₁ + 3r = 16

a₁ + 4r + a₁ + 6r = 28

2a₁ + 4r = 16 *(-1)

2a₁ + 10r = 28

- 2a₁ - 4r = -16

2a₁ + 10r = 28

6r = 12

r = 12/6

r = 2

2a₁ + 10r = 28

2a₁ + 10*(2) = 28

2a₁ + 20 = 28

2a₁ = 28 - 20

2a₁ = 8

a₁ = 8/2

a₁ = 4

a₂₈ = a₁ + 27r

a₂₈ = 4 + 27*(2)

a₂₈ = 4 + 54

a₂₈ = 58

Sn = (a₁ + an) * n / 2

S₂₈ = (a₁ + a₂₈) * 28 / 2

S₂₈ = (4 + 58) * 28 / 2

S₂₈ = (4 + 58) * 14

S₂₈ = 62 * 14

S₂₈ = 868