To solve the problem, we need to find the sum of all possible common differences of three natural numbers in an arithmetic progression (AP) that can be removed from the first 21 natural numbers such that the mean of the remaining numbers is 11.
### Step-by-Step Solution:
1. **Understand the Problem**:
We have the first 21 natural numbers: \(1, 2, 3, \ldots, 21\). We need to remove three numbers from this set that are in an arithmetic progression (AP). After removing these numbers, the mean of the remaining numbers should be 11.
2. **Calculate the Total Sum of First 21 Natural Numbers**:
The sum \(S\) of the first \(n\) natural numbers is given by the formula:
\[
S = \frac{n(n + 1)}{2}
\]
For \(n = 21\):
\[
S = \frac{21 \times 22}{2} = 231
\]
3. **Calculate the Sum of Remaining Numbers**:
Let the three numbers removed be \(a\), \(a+d\), and \(a+2d\) where \(d\) is the common difference. The sum of the three numbers removed is:
\[
\text{Sum of removed numbers} = a + (a + d) + (a + 2d) = 3a + 3d = 3(a + d)
\]
Thus, the sum of the remaining numbers is:
\[
\text{Sum of remaining numbers} = 231 - 3(a + d)
\]
4. **Set Up the Mean Equation**:
The mean of the remaining numbers is given as 11. The number of remaining numbers after removing 3 is \(21 - 3 = 18\).
Therefore, the mean can be expressed as:
\[
\frac{231 - 3(a + d)}{18} = 11
\]
5. **Solve for \(a + d\)**:
Multiplying both sides by 18:
\[
231 - 3(a + d) = 198
\]
Rearranging gives:
\[
3(a + d) = 231 - 198 = 33
\]
Thus:
\[
a + d = 11
\]
6. **Determine Possible Values for \(a\) and \(d\)**:
Since \(a\) must be a natural number and \(a + d = 11\), we can express \(d\) as:
\[
d = 11 - a
\]
The values of \(a\) can range from 1 to 10 (since \(d\) must also be a natural number).
7. **Calculate Possible Values of \(d\)**:
For \(a = 1\), \(d = 10\)
For \(a = 2\), \(d = 9\)
For \(a = 3\), \(d = 8\)
For \(a = 4\), \(d = 7\)
For \(a = 5\), \(d = 6\)
For \(a = 6\), \(d = 5\)
For \(a = 7\), \(d = 4\)
For \(a = 8\), \(d = 3\)
For \(a = 9\), \(d = 2\)
For \(a = 10\), \(d = 1\)
8. **Sum of All Possible Common Differences**:
The possible values of \(d\) are \(10, 9, 8, 7, 6, 5, 4, 3, 2, 1\). The sum of these values is:
\[
10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55
\]
### Final Answer:
The sum of all common differences of all such possible increasing AP is \(55\).
