Why is maximum extension always double the extension at mean position?

Assume an ideal spring of spring constant $$k$$ is connected with two blocks of masses $$m_1,m_2$$ at both of its end. And this system is kept on the horizontal smooth table. And a force $$F$$ is applied to $$m_1$$ in direction away from spring. And the task is to find maximum extension during journey.

So, I gone in the frame of $$m_1$$ assume acceleration of $$m_i$$ is $$a_i$$ and $$a$$ is acceleration of center of mass of the system. Then wrt $$m_1$$ acceleration of $$m_2$$ will be $$a_2-a_1$$. For the equilibrium position $$a_1=a_2=a$$

$$F-kx=m_1a$$$$kx=m_2a$$

(In ground frame)

Since this extension ($$x$$) is at mean position and the motion is SHM. So, maximum extension $$X=2x$$

Which on solving gives $$X=\frac{2m_2F}{(m_1+m_2)k}$$

Which I also verified by conservation of energy wrt Center of mass.

But I want to know why always it appears to be double the extension on mean position in case where constant force is applied?