(i) tn = 4n – 3 (1 mark)

(i) t_n = 4n – 3 (1 mark)

Sol. t_n = 4n – 3

∴ t₁ = 4 (1) – 3 = 4 – 3 = 1

∴ t₂ = 4 (2) – 3 = 8 – 3 = 5

∴ t₃ = 4 (3) – 3 = 12 – 3 = 9

∴ t₄ = 4 (4) – 3 = 16 – 3 = 13

∴ t₅ = 4 (5) – 3 = 20 – 3 = 17

∴ The first five terms of the sequence are 1, 5, 9, 13 and 17.