If the sum of first 11 terms of an `A.P., a_(1),a_(2),a_(3),....` is `0(a_(1)!=0)` then the sum of the `A.P., a_(1), a_(3),a_(5), ......,a_(23)` is `k a_(1)`, where k is equal to :

Options

A. `-(121)/(10)`
B. `-(72)/(5)`
C. `(72)/(5)`
D. `(121)/(10)`

Answer

To solve the problem, we need to find the value of \( k \) when the sum of the terms \( a_1, a_3, a_5, \ldots, a_{23} \) of an arithmetic progression (A.P.) is expressed as \( k a_1 \), given that the sum of the first 11 terms of the A.P. is 0.

### Step-by-Step Solution:

1. **Understanding the Sum of the First 11 Terms of A.P.**:
The sum of the first \( n \) terms of an A.P. can be calculated using the formula:
\[ S_n = \frac{n}{2} \left(2a_1 + (n-1)d\right) \]
For the first 11 terms, we have \( n = 11 \):
\[ S_{11} = \frac{11}{2} \left(2a_1 + 10d\right) = 0 \]

2. **Setting Up the Equation**:
Since \( S_{11} = 0 \):
\[ \frac{11}{2} (2a_1 + 10d) = 0 \]
This implies:
\[ 2a_1 + 10d = 0 \]
Rearranging gives:
\[ 2a_1 = -10d \quad \Rightarrow \quad a_1 = -5d \]

3. **Finding the Terms of the New A.P.**:
The new sequence we are interested in is \( a_1, a_3, a_5, \ldots, a_{23} \). This sequence consists of the odd-indexed terms of the original A.P.
The general term of the A.P. is given by:
\[ a_n = a_1 + (n-1)d \]
For odd terms:
\[ a_1, a_3, a_5, \ldots, a_{23} \]
The \( n \)-th odd term can be expressed as:
\[ a_{2n-1} = a_1 + (2n-2)d \]

4. **Determining the Number of Terms**:
The last term \( a_{23} \) corresponds to \( n = 12 \) (since \( 1, 3, 5, \ldots, 23 \) gives us 12 terms).

5. **Calculating the Sum of the New A.P.**:
The sum of the first 12 odd-indexed terms is:
\[ S_{12} = \frac{12}{2} \left(2a_1 + (12-1)(2d)\right) = 6 \left(2a_1 + 11 \cdot 2d\right) \]
Simplifying this:
\[ S_{12} = 6 \left(2a_1 + 22d\right) \]

6. **Substituting \( a_1 \)**:
We already found \( a_1 = -5d \):
\[ S_{12} = 6 \left(2(-5d) + 22d\right) = 6 \left(-10d + 22d\right) = 6 \cdot 12d = 72d \]

7. **Expressing in Terms of \( a_1 \)**:
Since \( a_1 = -5d \), we can express \( d \) in terms of \( a_1 \):
\[ d = -\frac{a_1}{5} \]
Substituting this into the sum:
\[ S_{12} = 72 \left(-\frac{a_1}{5}\right) = -\frac{72}{5} a_1 \]

8. **Finding \( k \)**:
From the expression \( S_{12} = k a_1 \), we have:
\[ k = -\frac{72}{5} \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{-\frac{72}{5}} \]