The main thing to note is that 53 will be the (n + 2) th term of this AP. Also, if all the terms in the series are integers then the common difference for this AP must also be an integer.

Problem: There are n arithmetic mean between 11 and 53. Each of them is an integer. How many distinct arithmetic progressions are possible from the above data?

Solution:

Step 1: Find the common difference
The first step is to find the common difference between the terms of the arithmetic progression. We know that the difference between the first term (11) and the last term (53) is 42. If there are n arithmetic means between them, then the common difference will be 42/(n+1).

Step 2: Find the possible values of n
Since the arithmetic means are integers, we know that the common difference must be an integer. This means that 42/(n+1) must be an integer. So, we need to find all the factors of 42 and see which ones will give us an integer value of (n+1).

The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. We can see that only the values (n+1) = 1, 2, 3, 6, and 7 will give us an integer value for the common difference.

Step 3: Find the number of arithmetic progressions for each value of n
For each possible value of (n+1), we need to find the number of distinct arithmetic progressions that can be formed. The formula for the number of arithmetic progressions with n terms is (n-2C2). Since we have (n+1) terms in each case, the number of arithmetic progressions will be (n-1)C2.

Using this formula, we can find the number of arithmetic progressions for each possible value of (n+1):

- When (n+1) = 1, there is only one term between 11 and 53, so there are no arithmetic progressions.
- When (n+1) = 2, there are two arithmetic means. The common difference is 14. There are 2 arithmetic progressions: 11, 25, 39, 53 and 53, 39, 25, 11.
- When (n+1) = 3, there are three arithmetic means. The common difference is 10. There are 3 arithmetic progressions: 11, 21, 31, 41, 53; 53, 43, 33, 23, 11; and 11, 23, 33, 43, 53.
- When (n+1) = 6, there are six arithmetic means. The common difference is 4. There are 15 arithmetic progressions.
- When (n+1) = 7, there are seven arithmetic means. The common difference is 3. There are 21 arithmetic progressions.

Step 4: Find the total number of distinct arithmetic progressions
Finally, we add up the number of arithmetic progressions for each value of (n+1) to get the total number of distinct arithmetic progressions:

0 + 2 + 3 + 15 + 21 = 41

Therefore, there are 41 distinct arithmetic progressions that can be formed using the given data.