**Solution:**

Given,

Second term, a _{2} = 14

Third term, a _{3} = 18

Common difference, d = a_{3 }− a_{2} = 18 − 14 = 4

a_{2} = a + d

14 = a + 4

a = 10 = First term

The sum of n terms;

S_{n} = n /2 [2a + (n – 1) d ]

S_{51} = 51/2 [2 × 10 (51 – 1) 4]

= 51/2 [20 + (50) × 4]

= 51 × 220/2

= 51 × 110

= 5610