The period of a physical pendulum is $T = 2\pi\sqrt{I/(mgh)}$, where $h$ is the distance from the pivot to the center of mass (CM), and $I$ is the rotational inertia about the pivot. For a meter stick of length $L_s=1.0$ m, the CM is at the 50 cm mark, so $h=d$. The rotational inertia about the CM is $I_{cm} = \frac{1}{12}mL_s^2$. Using the parallel-axis theorem, the inertia about the pivot is $I = I_{cm} + md^2 = \frac{1}{12}mL_s^2 + md^2$.

Substituting into the period formula:

$$T = 2\pi\sqrt{\frac{\frac{1}{12}mL_s^2 + md^2}{mgd}} = 2\pi\sqrt{\frac{L_s^2/12 + d^2}{gd}}$$

Squaring and rearranging gives a quadratic equation for $d$:

$$d^2 - \left( \frac{gT^2}{4\pi^2} \right)d + \frac{L_s^2}{12} = 0$$

Substituting $T = 2.5$ s, $L_s = 1.0$ m, and $g = 9.8$ m/s$^2$:

$$d^2 - \left( \frac{(9.8)(2.5)^2}{4\pi^2} \right)d + \frac{(1.0)^2}{12} = 0$$ $$d^2 - 1.551d + 0.08333 = 0$$

Using the quadratic formula, $d = \frac{1.551 \pm \sqrt{(-1.551)^2 - 4(1)(0.08333)}}{2}$. The smaller root corresponds to the pivot point being on the stick.