i. Given, t_n = 4n – 3

For n = 1, t₁ = 4(1) – 3 = 4 – 3 = 1

For n = 2, t₂ = 4(2) – 3 = 8 – 3 = 5

For n = 3, t₃ = 4(3) – 3 = 12 – 3 = 9

For n = 4, t₄ = 4(4) – 3 = 16 – 3 = 13

For n = 5, t₅ = 4(5) – 3 = 20 – 3 = 17 

\(\therefore\) The first five terms are 1, 5, 9, 13 and 17.

ii. Given, t_n = 2n – 5 For n = 1,

t₁ = 2(1) – 5 = –3

For n = 2, t₂ = 2(2) – 5 = –1

For n = 3, t₃ = 2(3) – 5 = 1

For n = 4, t₄ = 2(4) – 5 = 3

For n = 5, t₅ = 2(5) – 5 = 5

\(\therefore\) The first five terms are –3, –1, 1, 3 and 5.

iii. Given, t_n = n + 2

For n = 1, t₁ = 1 + 2 = 3

For n = 2, t₂ = 2 + 2 = 4

For n = 3, t₃ = 3 + 2 = 5

For n = 4, t₄ = 4 + 2 = 6

For n = 5, t₅ = 5 + 2 = 7

\(\therefore\) The first five terms are 3, 4, 5, 6 and 7.

iv. Given, t_n = n² – 2n

For n = 1, t₁ = (1)² – 2(1) = 1 – 2 = -1

For n = 2, t₂ = (2)² – 2(2) = 4 – 4 = 0

For n = 3, t₃ = (3)² – 2(3) = 9 – 6 = 3

For n = 4, t₄ = (4)² – 2(4) = 16 – 8 = 8

For n = 5, t₅ = (5)² – 2(5) = 25 – 10 = 15

\(\therefore\) The first five terms are –1, 0, 3, 8 and 15.

v. Given, t_n = n³

For n = 1, t₁ = (1)³ = 1

For n = 2, t₂ = (2)³ = 8

For n = 3, t₃ = (3)³ = 27

For n = 4, t₄ = (4)³ = 64

For n = 5, t₅ = (5)³ = 125

\(\therefore\) The first five terms are 1, 8, 27, 64 and 125.