1. Let the A.P. be $a_k=a_1+(k-1)d$ where $a_1\neq 0$ and $d$ is the common difference.

2. We are given $\sum_{k=1}^{12} a_{2k-1}=-\frac{72}{5}a_1.$

   Now, $a_{2k-1}=a_1+(2k-2)d=a_1+2(k-1)d.$

   So the terms $a_1,a_3,a_5,\dots,a_{23}$ themselves form an A.P. with first term $a_1$ and common difference $2d$.

3. Sum of first $12$ odd-positioned terms: $\sum_{k=1}^{12} a_{2k-1}=\frac{12}{2}\Big(2a_1+(12-1)(2d)\Big).$

   Hence, $\sum_{k=1}^{12} a_{2k-1}=6(2a_1+22d)=12a_1+132d.$

   Given, $12a_1+132d=-\frac{72}{5}a_1.$

4. Solve for $d$: $132d=-\frac{72}{5}a_1-12a_1$ $132d=-\frac{72+60}{5}a_1=-\frac{132}{5}a_1$ $d=-\frac{1}{5}a_1.$

5. Now use the condition $\sum_{k=1}^{n} a_k=0.$ For an A.P., $S_n=\frac{n}{2}\big(2a_1+(n-1)d\big).$

   Since $n\neq 0$, we need $2a_1+(n-1)d=0.$

   Substitute $d=-\frac{1}{5}a_1$: $2a_1+(n-1)\left(-\frac{1}{5}a_1\right)=0.$

   Because $a_1\neq 0$, divide by $a_1$: $2-\frac{n-1}{5}=0$ $\frac{n-1}{5}=2$ $n-1=10$ $n=11.$

6. Checking options:

   • A: $18$ ✗
   • B: $17$ ✗
   • C: $11$ ✓
   • D: $10$ ✗

Therefore, the correct answer is C: $11$.