shawndx wrote:
The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?
a) a1=-1, a2=-1.5
b) a1=-1, a2=1
c) a1=1, a2=-1
d) a1=1, a2=1.5
e) a1=1.5, a2=1
I am having trouble understanding what the question is asking and also solving it
Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.
Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater)
This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)
If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'
So what can you deduce about \(a_1\) and \(a_2\)?
1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)
2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)
Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it.
Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it.
Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.
Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).
OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D)
To be honest what is probability of such questions landing up on your gmat and on what level ...
I think, I would just flip the computer table and walk out rather than solve such types