Here, a₂ = 14, a₃ = 18 and n = 51.
a_n = a + (n – 1)d
⇒ a₂ = a + (2 -1)d
⇒ 14 = a + d
⇒ a = 14 – d …(1)
and a₃ = a + (3 -1)d
⇒ 18 = a + 2d
Putting the value of a from equation (1), we get
⇒ 18 = 14 – d + 2d
⇒ d = 4
Putting the value of d in equation (1), we get
⇒ a = 14 – 4 = 10
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S₅₁ = 51/2[2(10) +(51 -1) (4)]
⇒ S₅₁ = 51/2[220] = 5610.