Solutions:

(i) a_n = 3n + 2

Given sequence whose a_n = 3n + 2

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 and we get

a₁ = (3 × 1) + 2 = 3 + 2 = 5

a₂ = (3 × 2) + 2 = 6 + 2 = 8

a₃ = (3 × 3) + 2 = 9 + 2 = 11

a₄ = (3 × 4) + 2 = 12 + 2 = 14

a₅ = (3 × 5) + 2 = 15 + 2 = 17

∴ the required first five terms of the sequence whose n^th term, a_n = 3n + 2 are 5, 8, 11, 14, 17.

(ii) a_n = (n – 2)/3

Given sequence whose 

On putting n = 1, 2, 3, 4, 5 then can get the first five terms

∴ the required first five terms of the sequence whose n^th term,

(iii) a_n = 3ⁿ

Given sequence whose a_n = 3ⁿ

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above

a₁ = 3¹ = 3;

a₂ = 3² = 9;

a₃ = 27;

a₄ = 3⁴ = 81;

a₅ = 3⁵ = 243.

∴ the required first five terms of the sequence whose n^th term, a_n = 3ⁿ are 3, 9, 27, 81, 243.

(iv) a_n = (3n – 2)/ 5

Given sequence whose

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above

And, we get

∴ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5

(v) a_n = (-1)ⁿ2ⁿ

Given sequence whose a_n = (-1)ⁿ2ⁿ

To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

a₁ = (-1)¹.2¹ = (-1).2 = -2

a₂ = (-1)².2² = (-1).4 = 4

a₃ = (-1)³.2³ = (-1).8 = -8

a₄ = (-1)⁴.2⁴ = (-1).16 = 16

a₅ = (-1)⁵.2⁵ = (-1).32 = -32

∴ the first five terms of the sequence are – 2, 4, – 8, 16, – 32.

(vi) a_n = n(n – 2)/2

The given sequence is,

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And, we get

∴ the required first five terms are -1/2, 0, 3/2, 4, 15/2

(vii) a_n = n² – n + 1

The given sequence whose, a_n = n² – n + 1

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5.

And, we get

a₁ = 1² – 1 + 1 = 1

a₂ = 2² – 2 + 1 = 3

a₃ = 3² – 3 + 1 = 7

a₄ = 4² – 4 + 1 = 13

a₅ = 5² – 5 + 1 = 21

∴ the required first five terms of the sequence are 1, 3, 7, 13, 21.

(viii) a_n = 2n² – 3n + 1

The given sequence whose a_n = 2n² – 3n + 1

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And, we get

a₁ = 2.1² – 3.1 + 1 = 2 – 3 + 1 = 0

a₂ = 2.2² – 3.2 + 1 = 8 – 6 + 1 = 3

a₃ = 2.3² – 3.3 + 1 = 18 – 9 + 1 = 10

a₄ = 2.4² – 3.4 + 1 = 32 – 12 + 1 = 21

a₅ = 2.5² – 3.5 + 1 = 50 – 15 + 1 = 36

∴ the required first five terms of the sequence are 0, 3, 10, 21, 36.

(ix) a_n = (2n – 3)/ 6

Given sequence whose,

To get the first five terms of the sequence we put n = 1, 2, 3, 4, 5.

And, we get

∴ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6