The nth term for the series is a+(n-1)d where d is the common difference.
S1=a=6.
So S4=105=6+3d, 3d=105-6=99, d=33. The series is 6, 39, 72, 105, 138, 171, 204, ...
Sn=6+33(n-1)=6+33n-33=33n-27.
Sn-3=33(n-3)-27=33n-99-27=33n-126. Sn-3=Sn-99.
Sn/Sn-3=(33n-27)/(33n-126)=(11n-9)/(11n-42).
This is not (n+3)/(n-3) generally.
Let n=4, then S4/S1=105/6=35/2. But (n+3)/(n-3)=7.
If (11n-9)/(11n-42)=(n+3)/(n-3), then:
(11n-9)(n-3)=(11n-42)(n+3),
11n2-42n+27=11n2-9n-126,
27+126=42n-9n,
153=33n, n=153/33=51/11. So since n must be an integer, Sn:Sn-3≠(n+3):(n-3) for any integer n>3.