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

Given, a₃ = 16 (Eq 1)     and    a₇ - a₅ = 12 (Eq 2)

By using formula a_n = a + (n - 1)d on Eq 1 we get,

⇒ a₃ = a + (3 - 1)d = 16
⇒ a + 2d = 16
⇒ a = 16 - 2d.             (Eq 3)

By using formula a_n = a + (n - 1)d on Eq 2 we get,

⇒ a₇ - a₅ = 12
⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12
⇒ a + 6d - a - 4d = 12
⇒ 2d = 12 ⇒ d = 6.

∴ a = 16 - 2d = 16 - 2(6) = 16 - 12 = 4.

    a₂ = a + d = 4 + 6 = 10,    a₃ = a₂ + d = 10 + 6 = 16,
    a₄ = a₃ + d = 16 + 6 = 22.

Hence, the A.P. is 4, 10, 16, 22 ….