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DS : kplan

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shinewine
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DS : kplan [#permalink] DS : kplan   26 Aug 2006, 07:32
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A
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D
E

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please try this question...
explain ur answer



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Re: DS : kplan [#permalink] DS : kplan   26 Aug 2006, 09:37
(1) As each term after a1 is divisible by 5 and 4, it is divisible by 20. Thus there are an infinite number of terms that are divisible by 20. SUFF

(2) No information is given about terms after a6- NOT SUFF
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Re: DS : kplan [#permalink] DS : kplan   26 Aug 2006, 09:47
shinewine
please try this question...
explain ur answer


Consider (1):
Rewrite given statement as:

a(n) = (4*5) 5^(n-2)
= 20*5^(n-2)

for all n>2,this will have 20 as a factor. - SUFF

Consider (2)
a(5)/a(4) = 4 and a(6)/a(5) = 5
We cannot get a common ration here without more data. - INSUFF

Answer: A
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Re: DS : kplan [#permalink] DS : kplan   26 Aug 2006, 09:55
kevincan
(1) As each term after a1 is divisible by 5 and 4, it is divisible by 20. Thus there are an infinite number of terms that are divisible by 20. SUFF

(2) No information is given about terms after a6- NOT SUFF


I agree that statement 1 is sufficient, but with statement 2, how many consecutive terms in a series are sufficient enough to set up an equation?
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Re: DS : kplan [#permalink] DS : kplan   26 Aug 2006, 10:19
GMATT73
kevincan
(1) As each term after a1 is divisible by 5 and 4, it is divisible by 20. Thus there are an infinite number of terms that are divisible by 20. SUFF

(2) No information is given about terms after a6- NOT SUFF

I agree that statement 1 is sufficient, but with statement 2, how many consecutive terms in a series are sufficient enough to set up an equation?


Unless they give you a way to calculate, there's no way you can get a7.



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Re: DS : kplan [#permalink]
26 Aug 2006, 10:19
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