Solution:

Provided here that,

The second term, a₂ = 14

The third term, a₃ = 18

The common difference, d = a₃−a₂ = 18−14 = 4

a₂ = a+d

14 = a+4

a = 10 = First term

The sum of n terms;

S_n = n/2 [2a + (n – 1)d]

S₅₁ = 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 5610