Where should the support be placed so that the 1 m long arm is in balance if 50 g and 150 g

Where should the support be placed so that the 1 m long arm is in balance if 50 g and 150 g weights are suspended at its ends?

According to the leverage rule
F1 * l1 = F2 * l2,
where
F1 and F2 are the forces acting at the ends of the lever, and
l1 and l2 – the length of the arms of these forces;
l1 + l2 = 1m;
F1 = m1 * g and F2 = m2 * g,
where
g is the acceleration of gravity;
m1 * g * l1 = m2 * g * l2;
l1 / l2 = m2 / m1 = 150/50 = 3;
l1 = 3 * l2;
3 * l2 + l2 = 4 * l2 = 1m
l2 = 0.25m;
The support should be placed at a distance of 0.25m from the end, at which a load with a mass of 150g is suspended.