Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Given a2=14, a3=18∴d=18−14=4
By using nth term tn=a+(n−1)d we have
∴a2=a+d⇒14=a+4⇒a=10
Sn=n/2[2a+(n−1)d]
S51=51/2[2×10+(51−1)4]=51/2(20+200)
S51=51×110=5610