Three numbers whose sum is 15 are in an arithmetic progression. The product of the 1st two numbers is 2 more than the 3rd. Find the three numbers.
Designate the three numbers, in order as x, y and z.
x<y<z
xy = z +2 or xy - 2 = z
x + y + z = 15
x = 2, y = 5, z = 8
To find the three numbers, let's assume that the common difference in the arithmetic progression is 'd'.
Let's denote the first number as 'a', the second number as 'a + d', and the third number as 'a + 2d'.
According to the given information, the sum of these three numbers is 15:
a + (a + d) + (a + 2d) = 15
Simplifying the equation, we get:
3a + 3d = 15
Next, we are told that the product of the first two numbers is 2 more than the third number. Therefore, we can write the equation:
a * (a + d) = (a + 2d) + 2
Expanding and simplifying this equation, we have:
a^2 + ad = a + 2d + 2
Let's rearrange the equation and simplify:
a^2 + ad - a - 2d - 2 = 0
a^2 + (d - 1)a - (2d + 2) = 0
Now, we have two equations:
1) 3a + 3d = 15
2) a^2 + (d - 1)a - (2d + 2) = 0
We can solve this system of equations to find the values of 'a' and 'd', which will give us the three numbers in the arithmetic progression.