Let first term of AP = a

common difference = d

number of terms = n

Given that 2nd term of AP = 14

=> a + d = 14..........1

3rd term of AP = 18

=> a + 2d = 18........2

After solving equation 1 and 2, we get

a = 10 and d = 4

Given number of term n = 51

Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}

=> (51/2)* {2*10 + (51-1)*4}

=> (51/2)* (20 + 50*4)

=> (51/2)* (20 + 200)

=> (51/2)* 220

=> 51* 110

=>5610

So sum of first 51 terms of AP = 5610