Find the sum of first 51 terms of an A.P. whose 2^nd and 3^rd terms are 14 and 18 respectively.

Given,

⇒ a₂ = a + (2 - 1)d

⇒ 14 = a + d

⇒ a = 14 - d …….(i)

Also,

⇒ a₃ = a + (3 - 1)d

⇒ 18 = a + 2d

⇒ a = 18 - 2d …….(ii)

From (i) and (ii) we get,

⇒ 14 - d = 18 - 2d

⇒ -d + 2d = 18 - 14

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a = 14 - 4 = 10.

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{51}{2}[2 \times 10 + (51 - 1) \times 4] \\[1em] = \dfrac{51}{2} \times (20 + 200) \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, sum of first 51 terms = 5610.