i. Given, tn = 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, tn = 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, tn = 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, tn = 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, tn = 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.