Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively
Given that,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
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 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610
