There are n arithmetic means (were `n in N`) between 11 and 53 such that each of them is an integer. How many distinct arithmetic progressions are prossible from the above data ?

Options

A. 7
B. 8
C. 14
D. 16

Explanation

To solve the problem, we need to find how many distinct arithmetic progressions can be formed with `n` arithmetic means between the numbers 11 and 53, where `n` is a natural number.

### Step-by-Step Solution:

1. **Identify the terms of the arithmetic progression (AP)**:
- The first term \( a_1 = 11 \)
- The last term \( a_{n+2} = 53 \)
- The total number of terms in the AP is \( n + 2 \) (including the two endpoints).

2. **Use the formula for the nth term of an AP**:
- The nth term of an AP is given by:
\[ a_n = a + (n-1) \cdot d \]
- For our case, we have:
\[ 53 = 11 + (n + 1) \cdot d \]
- Rearranging gives:
\[ 53 - 11 = (n + 1) \cdot d \]
\[ 42 = (n + 1) \cdot d \]

3. **Express the common difference \( d \)**:
- From the equation \( 42 = (n + 1) \cdot d \), we can express \( d \) as:
\[ d = \frac{42}{n + 1} \]

4. **Determine the conditions for \( d \)**:
- Since \( d \) must be an integer, \( n + 1 \) must be a divisor of 42.

5. **Find the divisors of 42**:
- The divisors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42.

6. **Calculate possible values of \( n \)**:
- Since \( n + 1 \) must be one of the divisors, we can find \( n \) by subtracting 1 from each divisor:
- If \( n + 1 = 1 \) → \( n = 0 \) (not a natural number)
- If \( n + 1 = 2 \) → \( n = 1 \)
- If \( n + 1 = 3 \) → \( n = 2 \)
- If \( n + 1 = 6 \) → \( n = 5 \)
- If \( n + 1 = 7 \) → \( n = 6 \)
- If \( n + 1 = 14 \) → \( n = 13 \)
- If \( n + 1 = 21 \) → \( n = 20 \)
- If \( n + 1 = 42 \) → \( n = 41 \)

7. **Count the valid natural numbers**:
- The valid values of \( n \) are: 1, 2, 5, 6, 13, 20, 41.
- This gives us a total of 7 distinct values for \( n \).

8. **Conclusion**:
- Therefore, the number of distinct arithmetic progressions possible is **7**.