Question

Find three numbers in AP whose sum is 15 and whose product is 105.

HINT: Let the numbers be (a – d), a, (a + d).

Answer 1

Given: Let the three numbers be (a – d), a, (a + d); their sum is 15 and product is 105.

Step-wise calculation:

1. Sum: (a – d) + a + (a + d) = 3a = 15

⇒ a = `15/3`

⇒ a = 5

2. Product: (a – d) · a · (a + d) = a(a² – d²) = 105. 

Substitute a = 5: 5(25 – d²) = 105 

⇒ 25 – d² = 21

⇒ d² = 25 – 21

⇒ d² = 4 

⇒ d = ±2

3. For d = 2 the numbers are 3, 5, 7.

For d = –2 the numbers are 7, 5, 3 same set in reverse order.

The three numbers are 3, 5, 7 or 7, 5, 3.