Solutions:

(i) Provided that, d = 5, S₉ = 75

In AP, the sum of n terms is,

S_n = n/2 [2a +(n -1)d]

As a result, the sum of the first nine terms is;

S₉ = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

The n^th term can also be written as;

a_n = a+(n−1)d

a₉ = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(ii) Provided that, a = 2, d = 8, S_n = 90

In AP, the sum of n terms is,

S_n = n/2 [2a +(n -1)d]

90 = n/2 [2a +(n -1)d]

⇒ 180 = n(4+8n -8) = n(8n-4) = 8n²-4n

⇒ 8n²-4n –180 = 0

⇒ 2n²–n-45 = 0

⇒ 2n²-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

As a result, n = 5 (as n only be a positive integer)

∴ a₅ = 8+5×4 = 34