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Step 1: Understand the Problem
We have an arithmetic progression (A.P.) with first term $a_1$ and common difference $d$. The sum of its first 11 terms is given to be zero. We need to find the sum of the terms with odd indices, namely $a_1 + a_3 + a_5 + \dots + a_{23}$, in the form $k \, a_1$, and determine the value of $k$.
Step 2: Express the Sum of the First 11 Terms
The sum of the first $n$ terms of an A.P. is given by:
$$
S_n = \frac{n}{2} \bigl[2a_1 + (n-1)d \bigr].
$$
For $n = 11$, the sum is zero:
$$
S_{11} = \frac{11}{2} \bigl[2a_1 + 10d \bigr] = 0.
$$
Step 3: Find the Relationship Between $a_1$ and $d$
From
$$
\frac{11}{2} \bigl[2a_1 + 10d \bigr] = 0,
$$
we get
$$
2a_1 + 10d = 0 \quad \Longrightarrow \quad a_1 + 5d = 0 \quad \Longrightarrow \quad d = -\,\frac{a_1}{5}.
$$
Step 4: Identify the Terms to be Summed
We need to find the sum of $a_1, a_3, a_5, \dots, a_{23}$. These are the terms with odd indices. Notice that the indices 1, 3, 5, …, 23 form an arithmetic sequence of indices with common difference 2, and there are 12 such terms (since 1 to 23, in steps of 2, gives 12 terms).
Step 5: Sum the Odd-Indexed Terms
Use the fact that the sum of an arithmetic progression with $m$ terms can also be written as
$$
\text{Sum} = \frac{m}{2} \bigl[\text{(first term)} + \text{(last term)}\bigr].
$$
Here, $m = 12$. The first term among these is $a_1$, and the last term is $a_{23} = a_1 + 22d$. Thus,
$$
a_1 + a_3 + a_5 + \cdots + a_{23} \;=\; \frac{12}{2} \Bigl(a_1 + \bigl(a_1 + 22d\bigr)\Bigr) \;=\; 6 \bigl(2a_1 + 22d\bigr).
$$
Substitute $d = -\frac{a_1}{5}$:
$$
2a_1 + 22 \Bigl(-\frac{a_1}{5}\Bigr)
= 2a_1 \;-\; \frac{22}{5} a_1
= \frac{10a_1 - 22a_1}{5}
= -\,\frac{12a_1}{5}.
$$
Hence,
$$
a_1 + a_3 + a_5 + \cdots + a_{23}
= 6 \left(-\,\frac{12a_1}{5}\right)
= -\,\frac{72}{5}\,a_1.
$$
Step 6: Conclusion
Therefore, the sum of the odd-indexed terms is
$$
-\frac{72}{5}\,a_1.
$$
Hence, the value of $k$ in $k\, a_1$ is
$$
k = -\frac{72}{5}.
$$
