Sequences

[shankar.ashwin]

A sequence consists of 24 non zero integers. If each term in the sequence after the second is the product of the previous two term, how many terms in the sequence are negative?

(A) The third term in the seq in positive
(B) The fourth term is negative

[gmatclubmember]

A sequence consists of 24 non zero integers. If each term in the sequence after the second is the product of the previous two term, how many terms in the sequence are negative?

(A) The third term in the seq in positive
(B) The fourth term is negative

Lets say the seq is : a1, a2, a1a2, a1.a2^2, a1^2.a2^3,...
If third term is positive then either a1 and a2 are both positive or negative. So we cannot decide anything about the signs of further terms.
A->Insufficient

If the fourth term is negative that doesnt tell us about the sign of a2 (a1 is definitely negative though). So again INSUFFICIENT.

If we combine both we deduce that both a1 and a2 are negative.(from first - we know that both have same sign and from second we know that a1 is negative).
So both are sufficient.

Is OA C.

Cheers
Ami/-

[shankar.ashwin]

It's B Got it in Kaplan 800, supposed to be a tough DS.
Their explanation is kind of complicated, thought someone would come up with a easier way. Let me know if you want that soln, I can type it out.

[gmatclubmember]

It's B Got it in Kaplan 800, supposed to be a tough DS.
Their explanation is kind of complicated, thought someone would come up with a easier way. Let me know if you want that soln, I can type it out.

I still think it should be C. In fact if we take a groups of 3 terms then every third term would be positive and rest negative which will give us 16 negative terms and 8 positive terms .

Could you please post Kaplan reasoning ?

Cheers
Ami/-

[TimeforGMAT]

Answer is B. Fourth term can be negative only when first two terms are:

Case 1:Both negative
Case 2:First is -ve and second is +ve.

Lets expand case 1 scenario: (N=-ve, P=+)
N,N,P,N,N,P,N,N,P...You can see that there are two negatives for every three numbers. This means you will have a total of 16 negative numbers.
Now, for case 2 scenario:
N,P,N,N,P,N,N,P,N...You can notice again that there are two negatives for every three numbers. This means you will have a total of 16 negative numbers.

So, we can definitely say that there are 16 negative numbers when fourth number is negative.

Cheers.[/b]

[GMATGuruNY]

A sequence consists of 24 non zero integers. If each term in the sequence after the second is the product of the previous two term, how many terms in the sequence are negative?

(A) The third term in the seq in positive
(B) The fourth term is negative

The signs of the first 2 terms determine the signs of all the terms that follow.
There are only 4 options:
A: positive, positive, positive, positive, positive, positive.
B: positive, negative, negative, positive, negative, negative...
C: negative, negative, positive, negative, negative, positive...
D: negative, positive, negative, negative, positive, negative...

Statement 1: The 3rd term is positive.
In options A and C the 3rd term is positive.
In A, there are 0 negative terms.
In C, out of every 3 terms, two are negative.
Insufficient.

Statement 2: The 4th term is negative.
In options C and D the 4th term is negative.
In both, out of every 3 terms, two are negative.
Thus, the total number of negative terms = (2/3)*24 = 16.
Sufficient.

The correct answer is B.

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Yes correct there is only one case possible from the statement 2 that is being given...