Problem: The sum of three numbers in AP is 15 and the product of the first and the last is 21. Find the numbers.

Solution:

Let the three numbers be a-d, a, and a+d (since they are in AP).
Given, a-d + a + a+d = 15
=> 3a = 15
=> a = 5

Also, (a-d)(a+d) = 21
=> a² - d² = 21
Substituting a = 5, we get:
25 - d² = 21
=> d² = 4
=> d = ±2

Therefore, the three numbers are:
1. a-d = 5-2 = 3
2. a = 5
3. a+d = 5+2 = 7

Hence, the three numbers are 3, 5, and 7.

Explanation:

To solve this problem, we need to use the properties of arithmetic progression (AP) and algebraic equations. We can assume that the three numbers are in AP and represent them as a-d, a, and a+d. Then, we can use the given information to form equations and solve for the unknowns.

Firstly, we can use the fact that the sum of the three numbers is 15. This means that the sum of the first and last terms is twice the middle term, or 2a. Therefore, we can write:
a-d + a + a+d = 15
Simplifying this equation, we get:
3a = 15
a = 5

Now, we know the value of the middle term, a. Next, we can use the fact that the product of the first and last terms is 21. This means that the difference of their squares is 21. Therefore, we can write:
(a-d)(a+d) = 21
a² - d² = 21

Substituting a = 5, we can solve for d by simplifying the equation and taking the square root. This gives us two possible values for d: +2 and -2. Therefore, we get two sets of three numbers that are in AP: (3, 5, 7) and (7, 5, 3).

However, since the problem does not specify whether the AP is increasing or decreasing, both sets of numbers are valid solutions.