Given:
Four numbers are in arithmetic progression (A.P.).
The sum of the four numbers is 20.
The sum of the squares of the four numbers is 120.

To find:
Four numbers in A.P.

Approach:
Let's assume the four numbers in A.P. as a - 3d, a - d, a + d, and a + 3d. Here, 'a' represents the second term, and 'd' represents the common difference.

Using the formula for the sum of an arithmetic progression, we can write the equation:
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
Simplifying the equation, we get:
4a = 20
a = 5

Similarly, using the formula for the sum of the squares of an arithmetic progression, we can write the equation:
(a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 120
Expanding and simplifying the equation, we get:
12a^2 + 12d^2 = 120
a^2 + d^2 = 10

Solving for 'd':
Substituting the value of 'a' from the first equation into the second equation, we get:
25 + d^2 = 10
d^2 = -15

Since 'd^2' cannot be negative, there is no valid solution for 'd'. Therefore, there are no four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Conclusion:
There are no four numbers in A.P. that satisfy the given conditions.