If the sum of first 11 terms of an A.P. $$a_1, a_2, a_3, \ldots$$ is $$0$$ $$(a_1 \ne 0)$$ then the sum of the A.P. $$a_1, a_3, a_5, \ldots, a_{23}$$ is $$ka_1$$ where $$k$$ is equal to:

Let us denote the first term of the given arithmetic progression by $$a_1$$ and its common difference by $$d$$, so that the general term can be written as $$a_n = a_1 + (n-1)d$$.

First, we employ the standard formula for the sum of the first $$n$$ terms of an A.P.:

$$S_n = \frac{n}{2}\,[\,2a_1 + (n-1)d\,].$$

We are told that the sum of the first eleven terms is zero, i.e.

$$S_{11} = 0.$$

Substituting $$n = 11$$ into the sum formula, we obtain

$$0 \;=\; S_{11} \;=\; \frac{11}{2}\,\bigl[\,2a_1 + (11-1)d\,\bigr] \;=\; \frac{11}{2}\,\bigl[\,2a_1 + 10d\,\bigr].$$

Because neither $$\dfrac{11}{2}$$ nor $$a_1$$ is zero, the bracketed factor must vanish:

$$2a_1 + 10d = 0.$$

Dividing every term by $$2$$ gives

$$a_1 + 5d = 0 \quad\Longrightarrow\quad d = -\frac{a_1}{5}.$$

Now we need the sum of the subsequence $$a_1, a_3, a_5, \ldots, a_{23}.$$ Observe that these are the terms with odd indices, and the common difference between consecutive such terms is twice the original common difference, namely $$2d$$:

$$a_3 = a_1 + 2d,\; a_5 = a_1 + 4d,\;\ldots,\; a_{23} = a_1 + 22d.$$

To count how many terms are present, note that the indices run $$1,3,5,\ldots,23$$; this constitutes

$$\frac{23+1}{2} = 12$$

terms. Hence, within this new A.P. we have

• first term $$= a_1,$$
• common difference $$= 2d,$$
• number of terms $$= 12.$$

Applying the same sum formula with these values, we get

$$ S_{\text{odd}} \;=\; \frac{12}{2}\,\bigl[\,2a_1 + (12-1)(2d)\bigr] \;=\; 6\,\bigl[\,2a_1 + 11\cdot 2d\,\bigr] \;=\; 6\,\bigl[\,2a_1 + 22d\,\bigr]. $$

Expanding the bracket gives

$$S_{\text{odd}} = 12a_1 + 132d.$$

We already found $$d = -\dfrac{a_1}{5}.$$ Substituting this value of $$d$$ yields

$$ S_{\text{odd}} = 12a_1 + 132\left(-\frac{a_1}{5}\right) = 12a_1 - \frac{132}{5}a_1. $$

To combine the terms, express $$12a_1$$ with denominator $$5$$:

$$12a_1 = \frac{60}{5}a_1,$$

so

$$ S_{\text{odd}} = \frac{60}{5}a_1 - \frac{132}{5}a_1 = -\frac{72}{5}a_1. $$

Thus the required sum can be written as $$ka_1$$ with

$$k = -\frac{72}{5}.$$

Hence, the correct answer is Option D.