Let `A` be `nxxn` matrix given by
`A=[(a_(11),a_(12),a_(13)……a_(1n)),(a_(21),a_(22),a_(23)…a_(2n)),(vdots, vdots, vdots),(a_(n1),a_(n2),a_(n3).a_("nn"))]`
Such that each horizontal row is arithmetic progression and each vertical column is a geometrical progression. It is known that each column in geometric progression have the same common ratio. Given that `a_(24)=1,a_(42)=1/8` and `a_(43)=3/16`
The value of `lim_(nto oo)sum_(i=1)^(n)a_(ii)` is equal to

Options

A. `1/4`
B. `1/2`
C. `1`
D. `2`

Answer

To solve the problem step by step, we will analyze the given matrix properties and use the provided values to find the limit of the sum of the diagonal elements.

### Step 1: Understanding the Matrix Structure
The matrix \( A \) is defined such that:
- Each row is an arithmetic progression (AP).
- Each column is a geometric progression (GP) with the same common ratio \( r \).

### Step 2: Given Values
We are provided with the following values:
- \( a_{24} = 1 \)
- \( a_{42} = \frac{1}{8} \)
- \( a_{43} = \frac{3}{16} \)

### Step 3: Finding Common Difference in the Rows
Since \( a_{42} \) and \( a_{43} \) are part of an arithmetic progression, we can find the common difference \( d \):
\[ d = a_{43} - a_{42} = \frac{3}{16} - \frac{1}{8} = \frac{3}{16} - \frac{2}{16} = \frac{1}{16} \]

### Step 4: Finding More Elements in the Row
Using the common difference \( d \):
- \( a_{44} = a_{43} + d = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \)

### Step 5: Finding the Common Ratio in the Columns
We know that each column is a geometric progression. Let's denote the first element of the first column as \( a_{14} \). From the GP property:
\[ a_{24} = a_{14} \cdot r \quad \text{and} \quad a_{42} = a_{14} \cdot r^3 \]
Substituting the known values:
\[ 1 = a_{14} \cdot r \quad \text{and} \quad \frac{1}{8} = a_{14} \cdot r^3 \]

### Step 6: Finding the Common Ratio \( r \)
From \( a_{24} = 1 \):
\[ a_{14} = \frac{1}{r} \]
Substituting this into the equation for \( a_{42} \):
\[ \frac{1}{8} = \frac{1}{r} \cdot r^3 \implies \frac{1}{8} = \frac{r^2}{1} \implies r^2 = \frac{1}{8} \implies r = \frac{1}{2} \]

### Step 7: Finding \( a_{14} \)
Using \( r = \frac{1}{2} \):
\[ a_{14} = \frac{1}{r} = 2 \]

### Step 8: Finding Other Elements in the First Column
Using the common ratio \( r \):
- \( a_{24} = 1 \) (already known)
- \( a_{34} = a_{14} \cdot r^2 = 2 \cdot \left(\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2} \)
- \( a_{44} = a_{14} \cdot r^3 = 2 \cdot \left(\frac{1}{2}\right)^3 = 2 \cdot \frac{1}{8} = \frac{1}{4} \)

### Step 9: Finding the General Form of \( a_{ii} \)
The diagonal elements \( a_{ii} \) can be expressed as:
\[ a_{ii} = a_{i1} + (i-1) \cdot d = a_{14} \cdot r^{i-1} + (i-1) \cdot \frac{1}{2^i} \]
Substituting \( a_{14} = 2 \) and \( d = \frac{1}{2^i} \):
\[ a_{ii} = 2 \cdot \left(\frac{1}{2}\right)^{i-1} + (i-1) \cdot \frac{1}{2^i} = \frac{2}{2^{i-1}} + \frac{i-1}{2^i} \]

### Step 10: Finding the Limit of the Sum
We need to evaluate:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} a_{ii} \]
This becomes:
\[ \sum_{i=1}^{n} \left( \frac{2}{2^{i-1}} + \frac{i-1}{2^i} \right) \]
This can be split into two sums:
1. The first sum \( \sum_{i=1}^{n} \frac{2}{2^{i-1}} \) converges to \( 4 \) as \( n \to \infty \).
2. The second sum \( \sum_{i=1}^{n} \frac{i-1}{2^i} \) converges to \( 2 \).

Thus, the total limit is:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} a_{ii} = 4 + 2 = 6 \]

### Final Answer
The value of \( \lim_{n \to \infty} \sum_{i=1}^{n} a_{ii} \) is \( 6 \).