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Use the Chinese remainder theorem. Suppose we set 
. Then clearly taken modulo 12, the second term vanishes, and incidentally 
; taken modulo 13, the first term vanishes, but the second term leaves a remainder of 2. To counter this, we can multiply the second term by the inverse of 12 modulo 13, which is 12 since 
.
So, we found that
, but the least positive solution is 
, and in general we can have 
for any integer 
.
Now, since
, or 
, we know that there are 32 possible integers 
that satisfy the congruences.
So, we found that
Now, since
Apparently k=6. It doesn't say it in the question, but I'll assume that you had a mistype and forgot to say that the second term is 12.
Anyway,
The nth term of a geometric sequence is
. Thus, the 52nd term would be 
and the 50th term would be 
When you divide the two, the 2's would cancel out and k^49 would cancel out, leaving you with 1 on the bottom and k^2 on the top. As we said at the beginning, k=6, so our answer is just 6*6=36.
Anyway,
The nth term of a geometric sequence is