Given:

$ \sum\limits_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1 $

This means:

$ a_1 + a_3 + \cdots + a_{23} = -\frac{72}{5} a_1 $

Express each term in A.P.:

The general term of the A.P. is given by $ a_k = a + (k-1)d $. Thus, for the odd indices:

$ a_1 + (a + 2d) + (a + 4d) + \cdots + (a + 22d) $

Simplify the equation:

$ 12a + 2d(1 + 2 + \cdots + 11) = -\frac{72}{5} a $

Use the formula for sum of an arithmetic series (first 11 terms):

$ 1 + 2 + \cdots + 11 = \frac{11 \times 12}{2} = 66 $

So, the equation becomes:

$ 12a + 2d(66) = -\frac{72}{5} a $

$ 12a + 132d = -\frac{72}{5} a $

Solve for the relationship between $a$ and $d$:

$ 132d = -\frac{72}{5} a - 12a $

$ 132d = -\frac{132}{5} a $

So,

$ a = -5d \quad \text{(i)} $

Second condition:

$ \sum_{k=1}^{n} a_k = 0 \quad \Rightarrow \quad S_n = 0 $

$ \frac{n}{2} [2a + (n-1)d] = 0 $

$ 2a = -(n-1)d \quad \text{(ii)} $

Combine equation (i) and (ii):

Substitute $ a = -5d $ into equation (ii):

$ 2(-5d) = -(n-1)d $

$ -10d = -(n-1)d $

Therefore:

$ n-1 = 10 $

$ n = 11 $

Thus, $ n = 11 $.