To solve the problem, we will follow these steps:
### Step 1: Understand the Matrix Structure
Given that each row of the matrix \( A \) is an arithmetic progression (AP) and each column is a geometric progression (GP), we can denote the elements of the matrix \( A \) as follows:
- Let the first row be \( a_{11}, a_{12}, a_{13}, \ldots, a_{1n} \) where \( a_{1j} = a + (j-1)d \) for some constants \( a \) and \( d \).
- The second row will then be \( a_{21}, a_{22}, a_{23}, \ldots, a_{2n} \) which will also follow the same AP pattern but with a different starting point.
### Step 2: Use Given Values
We know:
- \( a_{24} = 1 \) (4th row, 2nd column)
- \( a_{42} = \frac{1}{8} \) (4th row, 2nd column)
- \( a_{43} = \frac{3}{16} \) (4th row, 3rd column)
### Step 3: Establish Relationships
From the information given, we can establish the following:
1. Since \( a_{24} = 1 \), we can denote this as \( a + 3d = 1 \).
2. For \( a_{42} = \frac{1}{8} \), we can denote this as \( a + 3d \cdot r = \frac{1}{8} \) where \( r \) is the common ratio of the GP.
3. For \( a_{43} = \frac{3}{16} \), we can denote this as \( a + 3d \cdot r^2 = \frac{3}{16} \).
### Step 4: Solve for \( d \) and \( r \)
From the equations:
1. \( a + 3d = 1 \) (1)
2. \( a + 3dr = \frac{1}{8} \) (2)
3. \( a + 3dr^2 = \frac{3}{16} \) (3)
From (1), we can express \( a \) in terms of \( d \):
\[ a = 1 - 3d \]
Substituting \( a \) into (2) and (3):
- From (2):
\[ 1 - 3d + 3dr = \frac{1}{8} \]
\[ 3dr - 3d = \frac{1}{8} - 1 \]
\[ 3d(r - 1) = -\frac{7}{8} \]
\[ d(r - 1) = -\frac{7}{24} \] (4)
- From (3):
\[ 1 - 3d + 3dr^2 = \frac{3}{16} \]
\[ 3dr^2 - 3d = \frac{3}{16} - 1 \]
\[ 3d(r^2 - 1) = -\frac{13}{16} \]
\[ d(r^2 - 1) = -\frac{13}{48} \] (5)
### Step 5: Find \( S_n \)
The sum \( S_n = \sum_{j=1}^{n} a_{4j} \) can be expressed as:
\[ S_n = n \cdot a + d \cdot \frac{n(n-1)}{2} \]
### Step 6: Compute Limit
We need to find:
\[ \lim_{n \to \infty} \frac{S_n}{n^2} \]
Substituting \( S_n \):
\[ S_n = n(1 - 3d) + d \cdot \frac{n(n-1)}{2} \]
Now, dividing by \( n^2 \):
\[ \frac{S_n}{n^2} = \frac{(1 - 3d)}{n} + \frac{d(n-1)}{2n} \]
Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{S_n}{n^2} = 0 + \frac{d}{2} = \frac{d}{2} \]
### Final Calculation
Using the values we found for \( d \) from the previous equations, we can substitute back to find the final limit.
### Conclusion
After simplifying, we find:
\[ \lim_{n \to \infty} \frac{S_n}{n^2} = \frac{1}{32} \]
