Let \( a_{1},a_{2},a_{3},\ldots \) be in an A.P. such that \( \displaystyle \sum_{k=1}^{12} 2a_{2k-1} = -\frac{72}{5} a_{1},\ a_{1} \ne 0 \). If \( \displaystyle \sum_{k=1}^{n} a_{k} = 0 \), then \( n \) is:
\( 17 \)
\( 10 \)
\( 18 \)
\( 11 \)