An < An-1 i.e. if the series is A1, A2, A3, A4..... then A4<A3....

St. 1) a25 = a24/2 or 2 (a25)=a24

Since a25 has to be less than a24, a24 has to be negative.

-insuff

St. 2) Prod. of 1st 25 terms is +ve implies that the no. of negative terms has to be an even no. - insuff.


I can't see how St. 1 & 2 together also shed any light on the no. of positive terms in the sequence.. I'd go with answer E but I might be missing out on something here.

ssgmatter wrote:In an infinite sequence of integers, the first term a1 = 40 and an < an-1 for all n
greater than 1.Are there more than 20 positive terms in this sequence ?
(1) a25 = a24/2
(2) The product of the first 25 terms of the sequence is positive.

Pick A

An < A(n-1)

Let assume that A(n-1) = -1. Then An = -0.5. This cannot be valid as -0.5> -1 whereas An < A(n-1).

So it gives to n understanding that A24 is a positive term.

So the previous terms are also positive.

So, Are there more than 20 positive terms in this sequence?? YES

Sufficient.

st 2 is Insufficient


Case 1: first 19 can be positive and the remaining 6 can be negative. But the ultimate product will be positive.

case 2: first 21 can be positive and remaining 4 can be negative. But the ultimate product will be positive.


So inconsistent.

Pick A

Good question almost fell for C.

But I go with A

If A25 is less than A24 and A25 = A24/2 A 24 has to be positive, whhich means all terms before it are also positive.

IMO A
s1) if a24 is +ve then a25 is also +ve and all an where n <24 is +ve so we have more than 20 +ves
if a24 is -ve a25 can not be less then a24 as a25 =a24/2 and a25 has to be less then a 24 so our case is not valid
suff
s2) insuff

The problem states "In an infinite sequence of integers, the first term a1 = 40 and an < an-1 for all n
greater than 1." How can there be -ve terms in the sequence?? an < an-1 holds good only for +ve numbers.

So as per stmt 2, if the product of first 25 terms is positive then can't we conclude that there are more than 20 terms in the sequence?
In that case ans should be D.
Can any one pls explain if this is wrong...

rash.patil wrote:The problem states "In an infinite sequence of integers, the first term a1 = 40 and an < an-1 for all n
greater than 1." How can there be -ve terms in the sequence?? an < an-1 holds good only for +ve numbers.

So as per stmt 2, if the product of first 25 terms is positive then can't we conclude that there are more than 20 terms in the sequence?
In that case ans should be D.
Can any one pls explain if this is wrong...

Hi rash,

You have slightly misread the question. The question says n is positive. But not An.

gmatmachoman wrote:
Pick A

An < A(n-1)

Let assume that A(n-1) = -1. Then An = -0.5. This cannot be valid as -0.5> -1 whereas An < A(n-1).

So it gives to n understanding that A24 is a positive term.

So the previous terms are also positive.

So, Are there more than 20 positive terms in this sequence?? YES

Sufficient.

Good question and very well explained by gmatmachoman..

sumanr84 wrote:

gmatmachoman wrote:
Pick A

An < A(n-1)

Let assume that A(n-1) = -1. Then An = -0.5. This cannot be valid as -0.5> -1 whereas An < A(n-1).

So it gives to n understanding that A24 is a positive term.

So the previous terms are also positive.

So, Are there more than 20 positive terms in this sequence?? YES

Sufficient.

Good question and very well explained by gmatmachoman..

Of course good explanation. But a small mistake..
An cannot be -0.5 as A1 to An are all integers.
Doesnt matter though