(ii) tn = 2n – 5 (1 mark)

(ii) t_n = 2n – 5 (1 mark)

Sol. t_n = 2n – 5

∴ t₁ = 2 (1) – 5 = 2 – 5 = – 3

∴ t₂ = 2 (2) – 5 = 4 – 5 = – 1

∴ t₃ = 2 (3) – 5 = 6 – 5 = 1

∴ t₄ = 2 (4) – 5 = 8 – 5 = 3

∴ t₅ = 2 (5) – 5 = 10 – 5 = 5

∴ The first five terms of the sequence are – 3, – 1, 1, 3 and 5.