Find the four numbers in A.P. whose sum is 50 and in which the greatest number is four times the least.

Let's solve this problem step by step. We are looking for four numbers in an arithmetic progression (A.P.) whose sum is 50, and the greatest number is four times the least number.

Let the four numbers be a, a + d, a + 2d, and a + 3d, where 'a' is the first term, and 'd' is the common difference of the arithmetic progression.

According to the problem, the greatest number is four times the least number, so we can write:

a + 3d = 4a

Now, we also know that the sum of the four numbers is 50. So, we can write the equation:

a + (a + d) + (a + 2d) + (a + 3d) = 50

Now, simplify this equation:

4a + 6d = 50

We have two equations:

a + 3d = 4a
4a + 6d = 50
We can solve this system of equations to find the values of 'a' and 'd'. Let's start with equation 1:

a + 3d = 4a

3d = 3a (subtracting 'a' from both sides)

d = a

Now that we know 'd' in terms of 'a', we can substitute this value into equation 2:

4a + 6(a) = 50

4a + 6a = 50

10a = 50

a = 50 / 10
a = 5

Now that we have found 'a', we can find 'd':

d = a = 5

So, the first term 'a' is 5, and the common difference 'd' is also 5.

Now, we can find the four numbers in the A.P.:

1st number: a = 5
2nd number: a + d = 5 + 5 = 10
3rd number: a + 2d = 5 + 2(5) = 15
4th number: a + 3d = 5 + 3(5) = 20

So, the four numbers in the arithmetic progression are 5, 10, 15, and 20, and their sum is indeed 50.