# What is the 6th term of the sequence? [closed]

Sequence: ABC; 1A1B1C; 111A111B111C; 311A311B311C; ...

What is the 6th term of the sequence?

a) 113121212A333311112B123112212C
b) 221133112A123212133B332211332C
c) 321321321A321321321B321321321C
d) 123123122A131221132B123123123C
e) 111312211A111312211B111312211C


How to solve this problem?

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## closed as off-topic by mrf, Trevor Wilson, dfeuer, Rick Decker, Vedran ŠegoOct 3 '13 at 0:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – mrf, Trevor Wilson, Vedran Šego
If this question can be reworded to fit the rules in the help center, please edit the question.

It seems to be a variant of the Look-and-say sequence. – Andrés Caicedo Oct 2 '13 at 22:51
(You begin with one A, one B, one C. The second term in the sequence tells you that this is what you had. This second term consists of one 1, one A, one 1, one B, one 1, and one C. Etc.) – Andrés Caicedo Oct 2 '13 at 22:52
The correct answer is c... ABC; 1A1B1C; 111A111B111C; 311A311B311C; 321A321B321C; 321321321A321321321B321321321C. – Victor Lellis Oct 2 '13 at 23:02
Victor, if you agree with the suggestion I'm giving of how the sequence is formed, your fifth and sixth terms are incorrect, the fourth terms begins with one 3, so the fifth term should begin 13... – Andrés Caicedo Oct 2 '13 at 23:09
Considering the variant of the Look-and-say sequence, the answer is e... but without this consideration and using another algorithm can be whatever answer. – Victor Lellis Oct 2 '13 at 23:30