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

Given that, Second term, a2 = 14 Third term, a3 = 18 Common difference, d = a3−a2 = 18−14 = 4 a2 = a+d 14 = a+4 a = 10 = First term Sum of n terms; 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