Odpowiedź:
zad 10
a₃ = a₁ + 2r = 17
a₄ + a₆ = a₁ + 3r + a₁ + 5r = 2a₁ + 8r = 46
układ równań
a₁ + 2r = 17 | * 2
2a₁ + 8r = 46
2a₁ + 4r = 34
2a₁ + 8r = 46
odejmujemy równania
2a₁ - 2a₁ + 4r - 8r = 34 - 46
- 4r = - 12
4r = 12
r = 12/4 = 3
a₁ + 2r = 17
a₁ + 2 * 3 = 17
a₁ + 6 = 17
a₁ = 17 - 6 = 11
an = a₁ + (n - 1) * r = 11 + (n - 1) * 3 = 11 + 3n - 3 = 3n + 8
zad 11
a₁ + a₂ + a₃ = a₁ + a₁ + r + a₁ + 2r = 3a₁ + 3r = 21
a₅ + 12 = a₂
a₁ + 4r = - 12 + a₁ + r
a₁ - a₁ + 4r - r = - 12
3r = - 12
r = - 12/3 = - 4
3a₁ + 3r = 21 | : 3
a₁ + r = 7
a₁ - 4 = 7
a₁ = 7 + 4 = 11
an = a₁ + (n - 1) * r = 11 + (n - 1) * (- 4) = 11 - 4n + 4 = - 4n + 15
