In an infinite sequence of integers, the first term a 1 = 40

Question

In an infinite sequence of integers, the first term a 1  = 40 and a n  is less than a n-1  for all n greater than 1.
Are there more than 20 positive terms in this sequence ?
(1) a 25  = a 24 /2
(2) The product of the first 25 terms of the sequence is positive.

trivikram · Answer

I will pick A here

if a25 = a24/2 
 then from the given description which is a(n)<a(n-1)

a(24) < 2*a(24) only if a(24) is +ve....

Since a(1) is +ve and a(25) is +ve and they being in a series the product of 1st 20 is +ve

From(B) I dont think we can know if some are -ve and some are +ve.
Not sure really here

But I would pick A

anonymousegmat · Answer

let me take a stab. i have been starting with statement two lately:

II - since we don't know the rule to build the sequence, this is insuff. we could have 24 negative numbers and 40.

I - i don't know if knowing the relationship between only two terms qualifies as the rule to build the entire sequence. we only know that 25th term is half the 24th term. i am tempted to say A - but other forces could be at work and the rule could be more complicated than each term being half the prior term.

I go with E.

FN · Answer

I am a lil confused...

if a1=40 then by definition An=An-1; so then a2=a2-1=a1??? which implies that a2=a1, which implies a3=a2=a1...based on that arent all the terms in the Sequence 40??

OK...good lord...re-read the question...it says An<An-1....big difference! 

OK now lets see

1)a25=a24/2 

we know that the first term is +, and the 25th term is half of 24...but we still dont know the sign of the 24th term...insuff

2) not much just says the entire term is positive..insuff

combining...not much either...E it is..

Kalyan · Answer

I'm going with A

1. a1=40 and a1<a0

a(25)=a(24)/2=a(23)/4=a(22)/8=.....=a(1)/2^19

So, all the terms from a(1) through a(25) are positive

2. Cant really do much with this information

bkk145 · Answer

(1) Term 25th can be greater or less than 1. Doesn't tell us anything about the rest. If it is less than 1, previous terms can be less than 1 as well. INSUFFICIENT

(2) This also doesn't tell much. We can have half negatives, half positives. or all positives. INSUFFICIENT.

I can't draw any conclusion from both. E.

bkk145 · Answer

I understand the OA=A, but from (1), if a(25) = a(24) /2, how can you assume that this is true for all cases of a(n)? It should have said:
a(n) = a(n-1) /2, then the answer will make sense.
Maybe I am missing something.

Robin in NC · Answer

A24 has to be a positive number. We know that A25<A24. If A25=A24/2, it must be positive. If it were negative, and were divided by 2, it would still be negative, and though it would have a lower absolute value, it would actually be a greater number. EX: if A24=-10, A25=-10/2=-5, which is greater, not less, than -10. So if n<n-1, and A25=A24/2, A24 has to be positive. So 1 is sufficient.

anonymousegmat · Answer

what is their rofficial explanation?

a1=40
a2 could be 39
a24 could be -2 or 2
a25 could be -1 or 1

without knowing the set builder notation how do we know?

trivikram · Answer

a25 = a24/2 
then from the given description which is a(n)<a(n-1) 
MEANS
a(24) < 2*a(24) only if a(24) is +ve

Hence it is A

Robin in NC · Answer

If a24 was -2, then a25 would be -1. But that violates the original conditions of the question stem - that An < An-1. A25 must be less than A24, and -1 is not less than -2. So a24 must be positive.

ywilfred · Answer

St1:
If a25 = -5, then a24 = -10. We are told a25 a24. So a25 cannot be positive, and there are thus more than 20 positive terms in this sequence.

St2:
We could have 14 positive terms and 11 negative terms and the product is positive. Or we could have all 25 terms positive, and the product is positive as well. Insufficient.

Ans A

In an infinite sequence of integers, the first term a 1 = 40

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