Understanding the Problem
To find three integers in Arithmetic Progression (AP) whose sum is 15 and product is 80, we can denote the integers as:
- a - d (first integer)
- a (second integer)
- a + d (third integer)
Here, 'a' is the middle term, and 'd' is the common difference.
Setting Up the Equations
1. Sum Equation:
The sum of the integers can be represented as:
(a - d) + a + (a + d) = 15
Simplifying this gives:
3a = 15
Thus, a = 5.
2. Product Equation:
The product of the integers is given by:
(a - d) * a * (a + d) = 80.
Substituting a = 5, we have:
(5 - d) * 5 * (5 + d) = 80.
Expanding and Solving
- Expanding the product yields:
(25 - d^2) * 5 = 80.
Thus, we have:
125 - 5d^2 = 80.
Rearranging gives:
5d^2 = 45, hence d^2 = 9.
- Therefore, d can be 3 or -3.
Finding the Integers
1. Using d = 3:
- First integer: 5 - 3 = 2
- Second integer: 5
- Third integer: 5 + 3 = 8
2. Using d = -3:
- First integer: 5 - (-3) = 8
- Second integer: 5
- Third integer: 5 + (-3) = 2
Conclusion
The three integers in AP are: 2, 5, and 8. Their sum (2 + 5 + 8) equals 15, and their product (2 * 5 * 8) equals 80.

2 , 5 and 8