AbhiJ wrote:D should be the answer.

Using GP rule

T(n) = a * r ^(n-1)
We know T(4), T(5), T(6), which gives the series.

Note: However we are not given with the above formula and the series can be some other series as well. If we were given the above formula then we could prove the above series using Principal of Mathematical Induction. However we are not given the formula hence GP for all n, cannot be proved.

Having said that this is out of the scope of GMAT Prep, and if this question appears on the GMAT the most likely answer would be D.

top_business_2011 wrote:
The answer is definitely A.
Let's see:

Required: Does the sequence contain an infinite number of terms divisible by 20?

Statement 1: Since al = 5, which is not divisible by 20, we can respond to the question saying,"No!" Hence, sufficient. [I think there is something wrong with statement 1, because even without 'an = 4(5^(n - 1)) for all integers n ≥ 2' the answer can be determined.]

Statement 2: We have no information about whether the sequence contains more than 6 terms.
the sequence may be just: { 20,100,500,2500, 12,500} In this case, the answer to our basic question is,"No!". However, if the sequence is continuous, then the answer is "Yes!". Hence, Insufficient.

Except a minor flow, in my opinion, in formulating statement 1, the problem is nice and is by no means outside the scope of the real GMAT. I have seen some questions,from trusted sources,testing similar concept.
Here is one of them: If P is a set of integers and 3 is in P, is every positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P. [Ans. A]

Whitney Garner wrote: Be careful, we only know that this is a GP from statement 1.

Statement (2) gives us NO information about the "formula" for this sequence - so it might not even have one. Therefore, we cannot know the values for any other terms except for those given, so this is Insufficient. Those values given ARE divisible by 20, but we cannot know about any of the others.

Extended FYI - this question seems fairly un-GMAT-like in that Statement (1) gives a1=5, a number clearly not divisible by 20, and still provides the geometric calculation for the rest of the series.


Whit

AbhiJ wrote:

top_business_2011 wrote:
The answer is definitely A.
Let's see:

Required: Does the sequence contain an infinite number of terms divisible by 20?

Statement 1: Since al = 5, which is not divisible by 20, we can respond to the question saying,"No!" Hence, sufficient. [I think there is something wrong with statement 1, because even without 'an = 4(5^(n - 1)) for all integers n ≥ 2' the answer can be determined.]

Statement 2: We have no information about whether the sequence contains more than 6 terms.
the sequence may be just: { 20,100,500,2500, 12,500} In this case, the answer to our basic question is,"No!". However, if the sequence is continuous, then the answer is "Yes!". Hence, Insufficient.

Except a minor flow, in my opinion, in formulating statement 1, the problem is nice and is by no means outside the scope of the real GMAT. I have seen some questions,from trusted sources,testing similar concept.
Here is one of them: If P is a set of integers and 3 is in P, is every positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P. [Ans. A]

I see your point that the series can be only 5 terms[If that's so, then it is NOT infinite]. But the question stem talks about a continous series (a1, a2, a3.....an...)?[ Well, here I agree with you that the question is somewhat ambiguous] Moreover in option B we are not said that it doesnot contain infinite terms. So one is free to assume that its a continous series by common logic.[Never restrict your domain, unless you're expressly told to do so! Just as we shouldn't assume a number to be an integer only because it is easier to deal with] Its the type of ambiguity you will not find in any OG problem.[ You can say that again!]