Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

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

By formula,

a_n = a + (n - 1)d

Given,

⇒ 3rd term = 16

⇒ a₃ = 16

⇒ a + (3 - 1)d = 16

⇒ a + 2d = 16 ……….(1)

Given,

7th term exceeds the 5th term by 12.

⇒ a₇ - a₅ = 12

⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12

⇒ a + 6d - [a + 4d] = 12

⇒ a - a + 6d - 4d = 12

⇒ 2d = 12

⇒ d = $\dfrac{12}{2}$ = 6.

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

⇒ a + 2 × 6 = 16

⇒ a + 12 = 16

⇒ a = 16 - 12 = 4.

A.P. : a, a + d, a + 2d, a + 3d, ……..

⇒ 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, …..

⇒ 4, 10, 16, 22, ……..

Hence, the required A.P. is 4, 10, 16, 22, ……..