Think outside the discrete

If the AP has starting term $a$ and common difference $d$ , then $S_n = \tfrac12n\big[2a + (n-1)d\big]$ . Since $S_6 > S_7$ , we deduce that $a + 6d < 0$ . Since $S_7 > S_5$ , we deduce that $2a + 11d > 0$ . Hence it follows that $d < 0$ and $\tfrac{11}{2}|d| < a < 6|d|$ . But then $S_{12} \; = \; 6(2a - 11|d|) > 0 > \tfrac{13}{2}(2a-12|d|) = S_{13}$ so that $S_{12}S_{13} < 0$ , making the answer $\boxed{12}$ .