Let the first term of arithmetic progression be A and the tolerance be D, and you can calculate.
S3=3a+3d
S6=6a+ 15d
S 12= 12a+66d
S3/S6= 1/3
Then (3a+3d)/(6a+15d) =1/3.
The solution is a=2d.
So S6/S 12=3/ 10.
===================================================
Arithmetic progression S3, S6-S3, S9-S6 and S 12-S9 also became arithmetic progression.
S3/S6= 1/3,S6=3S3,S6-S3=2S3
S9-S6=3S3,S9=6S3
S 12-S9=4S3,S 12= 10S3
So S6/S 12=3/ 10.
2.Sn is the sum of the first n terms of arithmetic progression {an}. It is known that S9= 18, Sn=240 and a(n-4)=30(n is greater than 9). How to find n?
Let the first term a, tolerance d.
(a+a+8d)*9/2= 18
[a+a+(n- 1)d]*n/2=240
a+(n-5)d=30
Hardware decoding, a =-50/3, d = 14/3, n = 15.
There is a better way:
(a+a+8d)*9/2= 18→a+4d=2
a+(n-5)d=30
Adding the two formulas, we can get 2a+nd-d=32.
Comparison [a+a+(n-1) d] * n/2 = 240 → (2a+nd-d) * n = 480.
Divided by n=480/32= 15.
Note: In fact, S9= 18, which means a5= 18/9=2.