Two weights, m and 2m, are suspended from a spring. How will the tension

Two weights, m and 2m, are suspended from a spring. How will the tension of the spring change if the weight of a larger mass is unhooked from it?

The force of elasticity balances the force of gravity acting on it from loads m and 2m:
Fcont. = Fт.
Fcont. = k * ∆l, where k is the stiffness of the spring; ∆l is the amount of spring deformation (elongation).
Fт = m * g, where m is the mass of loads (m, 2m), g is the acceleration of gravity.
Spring rate before uncoupling the load 2m:
∆l1 = 3m * g / k.
Spring stiffness after uncoupling the load 2m:
∆l2 = m * g / k.
∆l2 / ∆l1 = (m * g / k) / (3m * g / k) = 1/3.
Answer: b) The tension of the spring will decrease by 3 times.