Second term of A.P. a2 = 14
and third term a3 = 18
∴ Common difference d = a3 – a2
= 18 – 14 = 4
Again, ∵ IInd term = 14
∴ a + d = 14
⇒ a + 4 = 14
⇒ a = 14 – 4
⇒ a = 10
∵ a = 10, d = 4
Then, sum of n terms
Sn = n/2[2a + (n – 1)d]
∴ S51 = 51/2[2 × 10 + (51 – 1)4]
= 51/2[20 + 50 × 4]
= 51/2[20 + 200]
= 51/2 × 220 = 51 × 110 = 5610
Hence, sum of 5 terms of given A.P. = 5610