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

Given, a₂ = 14 and a₃ = 18.

common difference = d = any term - preceding term = a₃ - a₂ = 18 - 14 = 4.

By formula, a_n = a + (n - 1)d
⇒ a₂ = a + 4(2 - 1)
⇒ 14 = a + 4
⇒ a = 10.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{51} = \dfrac{51}{2}[2 \times 10 + (51 - 1)4] \\[1em] = \dfrac{51}{2}[20 + 50 \times 4] \\[1em] = \dfrac{51}{2}[20 + 200] \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, the sum of first 51 terms of the A.P. is 5610.