Math Wizards please help me with these number sequences ?

Question

1. __, 1, 5, 2, 8, 3, 5, __, 4, 5, 1, 4

2. 12, 57, 31, 14, 51, 138, 16, 71, 18, __, __, __

3. 34, 22, 89, 63, __,41, 12, __, 64, 92, 25, __

4. 31517, 114, 9111113, 148, __, __

5. 11534, 11343, 26828, 821135, __

Thanks in advance.

Colin

Answer 1

Ok i spent over an hour looking at these and i only came up with the answer for #2. Im not 100% sure, but this is as close as i can get. I found out it has 3 patterns:

1. The Sequence is A(1),X(1),A(2),A(3),
B(1),X(2),B(2),B(3),
C(1),X(3),C(2),C(3).
2. The way to determine the ABCs goes 1(Y), (Y)1,1(Y),(Y)1,
1(Y),(Y)1,1(Y),(Y)1,1(Y). Y being consecutive numbers from
2-10. Example, 1(2),(3)1,1(4),(5)1,1(6),(7)1,
1(8),(9)1,(10)1. Notice how the numbers are 2-10 but
with a 1 switchin back and forth. It goes left, right ,left, right, etc... those are the A B Cs.
3. Now, the answer for X. X is the result of A(1)+A(2)+A(3), B(1)+B(2)+B(3), and C(1)+C(2)+C(3). For example, X(1) is 57, and the A's are 12,31,and 14. 12+31+14 = 57. X(2) is 138, and the B's are 51,16,and 71. 51+16+71 = 138.

The third X is not given, neither is C(2) and C(3). So we follow the pattern. C(2) should be 91, since C(1) was 18, and according to pattern #1, 9 comes after 8, and the one has to be switched. And C(3) should be 110, since 10 comes after 9, and the one has to be switched back to the left. That leaves us with X(3). 18+91+110 = 219, therefore the last X is 219.

The final answer for #2 should look like this:
12,57,31,14,51,138,16,71,18,_,_,_.
(219) (91) (110)
Now i dont know what took longer, to solve it, or to explain how i did it haha xD. I've never seen this problem or anything like it, but i gotta admit i really love math.
-Let me know if you know for a fact that the answer is wrong.
-Can i know what this is for and what level of math this is? o.o

(I shall continue to try to solve the rest)

Answer 2

2) 129 ,91, 20

12+31+14=57
51+16+71=138
18+91+20=129

3) 23, 24, 73
89=43+22+(24)
63=23+14+(26)
64=12+24+(28)
92=25+a+(30). a=37
4) 31517, 114 (7*2=14)
9111113, 148 ( (1+3)*2=8; (1+1)*2=4)
15117119, 121620 ( (1+5)*2=12, (1+7)*2=16, (1+9)*2=20)

Answer 3

the fourth set of numbers do not make sense. are you able to recheck them. the only sequence I see is an straightforward multiplication-sum sequence one million x 6 = 6, 6 + 2 = 8, 2 x 6 = 12 ::: 06812 the subsequent type in the sequence would be 01484 and then 053211, so it is going to not be it.