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

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

2nd term = 14

⇒ a₂ = 14

⇒ a + (2 - 1)d = 14

⇒ a + d = 14 …………(1)

3rd term = 18

⇒ a₃ = 18

⇒ a + (3 - 1)d = 18

⇒ a + 2d = 18 …………(2)

Subtracting equation (1) from (2), we get :

⇒ a + 2d - (a + d) = 18 - 14

⇒ a - a + 2d - d = 4

⇒ d = 4.

Substituting value of d in equation (1), we get :

⇒ a + 4 = 14

⇒ a = 10.

Last term (l) = 51st term = a₅₁

= a + (51 - 1)d

= 10 + 50 × 4

= 10 + 200

= 210.

By formula,

Sum (S) = $\dfrac{n}{2}[a + l]$

Substituting values we get :

$\text{Sum of first 51 terms} = \dfrac{51}{2}[10 + 210] \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, sum of first 51 terms = 5610.