Question

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

Answer 1

A2 = 14 and a3 = 18
Common difference = a3 - a2 = 18 - 14 = 4 = d
Now
a2 = a+d=14
a+4=14
a = 10
Now, sum of 51 terms
={51(2a+(50)d)}/2
={51(20+200)}/2
={51×220}/2
=51×110=5610
Therefore sum of 51 terms is 5610