If four numbers. in decreasing A.P, a, b, c, d have their sum as 20 & sum of squares is 120 then a-b+c-d is

Explanation

To solve the problem, we need to find the values of \( a, b, c, d \) which are in decreasing arithmetic progression (A.P.) and satisfy the given conditions.

1. **Let the four numbers in decreasing A.P. be represented as:**
\[ a = A - 3d, \quad b = A - d, \quad c = A + d, \quad d = A + 3d \]
where \( A \) is the middle value and \( d \) is the common difference.

2. **Given the conditions:**
- The sum of the numbers is:
\[ a + b + c + d = 20 \]
- The sum of the squares of the numbers is:
\[ a^2 + b^2 + c^2 + d^2 = 120 \]

3. **Substituting the expressions for \( a, b, c, d \) into the sum:**
\[ (A - 3d) + (A - d) + (A + d) + (A + 3d) = 20 \]
Simplifying this:
\[ 4A = 20 \implies A = 5 \]

4. **Now substituting \( A = 5 \) into the expressions for \( a, b, c, d \):**
\[ a = 5 - 3d, \quad b = 5 - d, \quad c = 5 + d, \quad d = 5 + 3d \]

5. **Next, we substitute these into the sum of squares:**
\[ (5 - 3d)^2 + (5 - d)^2 + (5 + d)^2 + (5 + 3d)^2 = 120 \]

6. **Expanding each term:**
\[ (5 - 3d)^2 = 25 - 30d + 9d^2 \]
\[ (5 - d)^2 = 25 - 10d + d^2 \]
\[ (5 + d)^2 = 25 + 10d + d^2 \]
\[ (5 + 3d)^2 = 25 + 30d + 9d^2 \]

7. **Combining all these:**
\[ (25 - 30d + 9d^2) + (25 - 10d + d^2) + (25 + 10d + d^2) + (25 + 30d + 9d^2) = 120 \]
Simplifying gives:
\[ 100 + 20d^2 = 120 \]

8. **Solving for \( d^2 \):**
\[ 20d^2 = 20 \implies d^2 = 1 \implies d = \pm 1 \]

9. **Since \( a, b, c, d \) are in decreasing order, we take \( d = -1 \):**
\[ a = 5 - 3(-1) = 8, \quad b = 5 - (-1) = 6, \quad c = 5 + (-1) = 4, \quad d = 5 + 3(-1) = 2 \]

10. **Now we find \( a - b + c - d \):**
\[ a - b + c - d = 8 - 6 + 4 - 2 = 2 + 2 = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]