Let [tex]a_n[/tex] denote the n-th term in the progression. So
[tex]a_n=a_{n-1}+d[/tex]
for some constant difference between terms d.
Solve for [tex]a_n[/tex] explicitly:
[tex]a_4=a_3+d[/tex]
[tex]a_5=a_4+d=a_3+2d[/tex]
[tex]a_6=a_5+d=a_3+3d[/tex]
and so on, up to
[tex]a_n=a_3+(n-3)d[/tex]
We're told that the third term is [tex]a_3=4m-2n[/tex], and the ninth term is [tex]a_9=2m-8n[/tex], and according to the recursive rule above, we have
[tex]a_9=a_3+6d[/tex]
Solve for d :
[tex]2m-8n=(4m-2n)+6d[/tex]
[tex]-2m-6n=6d[/tex]
[tex]d=-\dfrac{2m+6n}6=\boxed{-\dfrac{m+3n}3}[/tex]
