Question

find the three no in arithmetic progression whose sum is 3 and whose product is - 15

Answer 1

Let the 3 terms of A.P be:

• (a - d)
• a
• (a + d)

In the given question,

☞ The sum of these 3 terms is 3.

(a - d) + a + (a + d) = 3

⇒ a - d + a + a + d = 3

⇒ 3a = 3

⇒ a = 3 ÷ 3

⇒ a = 1

So, The First term = a = 1

Also,

☞ Product of the 3 terms = -15

(a - d) a (a + d) = -15

⇒ (a - d) (a + d) a = -15

⇒ a (a² - d²) = -15

⇒ a³ - ad² = -15

⇒ 1³ - 1(d²) = -15

⇒ 1 - d² = -15

⇒ d² = 16

⇒ d = √16

⇒ d = ± 4

So, Common difference = d = ± 4

Case 1: If d = + 4

The 3 terms of A.P →

• (a - d) = 1 - 4 = -3
• a = 1
• (a + d) = 1 + 4 = 5

∴ The First 3 terms = -3, 1, 5

OR

Case 2: If d = -4

The 3 terms of the A.P →

• (a - d) = 1 - (-4) = 1 + 4 = 5
• a = 1
• (a + d) = 1 - 4 = -3

∴ The First 3 terms = 5, 1, -3