Let the three numbers in Arithmetic Progression (AP) be:
a – d, a, a + d
where a is the middle term and d is the common difference.
According to the problem, the sum of these three numbers is 18:
(a – d) + a + (a + d) = 18
Simplifying gives:
3a = 18
Thus, we find:
a = 6
Now, substituting a into the numbers, we have:
6 – d, 6, 6 + d
Next, we also know that the product of the first and third numbers is 5 times the common difference:
(6 – d)(6 + d) = 5d
Expanding the left side:
36 – d^2 = 5d
Rearranging the equation gives:
d^2 + 5d – 36 = 0
We will now solve this quadratic equation using the quadratic formula, d = [-b ± √(b² – 4ac)] / 2a, where a = 1, b = 5, c = -36.
Calculating the discriminant:
b² – 4ac = 5² – 4(1)(-36) = 25 + 144 = 169
Now substituting back, we find:
d = [-5 ± √169] / 2
d = [-5 ± 13] / 2
Calculating the roots:
d = (8 / 2) = 4 or d = (-18 / 2) = -9
Using these values of d, we find the numbers:
For d = 4:
6 – 4 = 2, 6, 6 + 4 = 10
The numbers are: 2, 6, 10.
For d = -9:
6 – (-9) = 15, 6, 6 + (-9) = -3
The numbers are: 15, 6, -3.
Thus, the possible sets of numbers that satisfy the conditions given in the problem are:
2, 6, 10 and 15, 6, -3.