Solutions:

(i) Provided that, a = 5, d = 3, a_n = 50

The formula for the nth term in an AP, as we know, is

a_n = a +(n −1)d,

Putting the given values, as a result we get,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

Now, the sum of n terms,

S_n = n/2 (a +a_n)

S_n = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a₁₃ = 35

The formula for the nth term in an AP, as we know, is

a_n = a+(n−1)d,

Putting the given values, as a result we get,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, S_n = n/2 (a+a_n)

S₁₃ = 13/2 (7+35) = 273