Question

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

Answer 1

Let the terms of the A.P be:

• (a - d)
• (a)
• (a + d)

According to the question,

Sum of first 3 terms = 33

⇒ (a - d) + a + (a + d) = 33

⇒ 3a = 33

⇒ a = 33 ÷ 3

∴ a = 11

The product of the first and third terms exceeds the Second term by 29

⇒ (a - d)(a + d) = a + 29

⇒ a² - d² = a + 29

→ 11² - d² = 11 + 29

→ 121 - d² = 40

→ d² = 121 - 40

→ d² = 81

→ d = √81

∴ d = ±9

★ Case 1: If d = +9

A.P ➔ (a - d), (a), (a + d), ...

⇒ A.P ➔ (11 - 9), (11), (11 + 9)

⇒ A.P ➔ 2, 11, 20...

★ Case 2: If d = -9

A.P ➔ (a - d), (a), (a + d)

⇒ A.P ➔ (11 + 9), (11), (11 - 9)

⇒ A.P ➔ 20, 11, 2...

∴ The A.P will be 2, 11, 20...

or 20, 11, 2...