The 40^th term of an A.P. exceeds its 16^th term by 72. Then its common difference is :

1. 40

2. 40 - 16

3. 72

4. 3

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

By formula,

a_n = a + (n - 1)d

Given,

40^th term of an A.P. exceeds its 16^th term by 72.

∴ a₄₀ - a₁₆ = 72

⇒ [a + (40 - 1)d] - [a + (16 - 1)d] = 72

⇒ a + 39d - [a + 15d] = 72

⇒ a - a + 39d - 15d = 72

⇒ 24d = 72

⇒ d = $\dfrac{72}{24}$ = 3.

Hence, Option 4 is the correct option.