The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.

Let the three numbers be a - d, a, a + d.

Given , sum of these terms = 33
⇒ a - d + a + a + d = 33
⇒ 3a = 33
⇒ a = 11.

Given, the product of the first and the third terms exceeds the second term by 29
⇒ (a + d)(a - d) - a = 29

Putting the value of a = 11 we get,

⇒ (11 + d)(11 - d) - 11 = 29
⇒ (11)² - d² - 11 = 29
⇒ 121 - d² - 11 = 29
⇒ 110 - d² = 29
⇒ 110 - 29 = d²
⇒ d² = 81
⇒ d = $\sqrt{81}$
⇒ d = +9, -9.

∴ a - d = 11 - 9 = 2 or 11 - (-9) = 20 and a + d = 11 + 9 = 20 or 11 + (-9) = 2.

Hence, the A.P. is 2, 11, 20, … or 20, 11, 2, ….