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

Solution:

Given,

Second term, a ₂ = 14

Third term, a ₃ = 18

Common difference, d = a₃ − a₂ = 18 − 14 = 4

a₂ = a + d

14 = a + 4

a = 10 = First term

The sum of n terms;

S_n = n /2 [2a + (n – 1) d ]

S₅₁ = 51/2 [2 × 10 (51 – 1) 4]

= 51/2 [20 + (50) × 4]

= 51 × 220/2

= 51 × 110

= 5610