Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Re: In a sequence of terms in which each term is three times the
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14 Jan 2016, 18:23
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
In a sequence of terms in which each term is three times the previous term, what is the fourth term?
(1) The first term is 3.
(2) The second-to-last term is 3^10.
Modify the original condition and the question and suppose the sequence A_n. Then it becomes A_(n+1)=3A_(n) and once you figure out A_(1), you can figure out everything. So there is 1 variable, which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), A_(1)=3 -> A_(2)=3^2, A_(3)=3^3, A_(4)=3^4, which is unique and sufficient.
For 2), you can’t figure out where the last term comes, which is not sufficient. Therefore, the answer is A.
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.