Understanding the Problem
To find three integers in arithmetic progression (AP) whose sum is 15 and product is 80, we can represent them as:
- Let the three integers be: a - d, a, a + d.
Setting Up the Equations
- Their sum:
- (a - d) + a + (a + d) = 15
- Simplifying gives: 3a = 15
- Thus, a = 5.
- Their product:
- (a - d) * a * (a + d) = 80
- Substituting a = 5 gives: (5 - d) * 5 * (5 + d) = 80.
Simplifying the Product
- Expanding the product:
- (5 - d)(5 + d) = 25 - d^2.
- Therefore, the equation becomes: 5(25 - d^2) = 80.
- Simplifying gives: 125 - 5d^2 = 80.
- Rearranging leads to: 5d^2 = 45, hence d^2 = 9.
Finding the Values of d
- Taking the square root gives:
- d = 3 or d = -3.
Calculating the Integers
- Using d = 3:
- The integers are: 5 - 3 = 2, 5, and 5 + 3 = 8.
- Using d = -3:
- The integers remain the same: 2, 5, and 8.
Final Integers
- The integers in AP that satisfy both the sum and product conditions are:
- 2, 5, and 8.
Thus, the three integers are 2, 5, and 8, fulfilling the given requirements.