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Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) +

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ikaytas
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Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   08 Jun 2006, 10:23
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Consider the series : a(1), a(2), ....a(15).
a(n) = a(n-1) + k, 2<=n<=15 and k is a non zero constant.
How many terms in the series are less than 10?

(1) a(1)= 24
(2) a(8)= 10

Answer is (b). I think it should be (c). Can someone pls explain why b.



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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   08 Jun 2006, 11:23
Took a quick look and C seems obvious. We dont know value of K so not sure how B. will take a closer look later
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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   08 Jun 2006, 11:34
I get C as well...not sure why the answer's B either since we don't know the value of K nor the value of a(1). Anyone got an explanation?
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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   08 Jun 2006, 16:19
it is b.

a[8] is the median. either a[1] thru a[7] are less than 10 or a[9] thru a[15] are less than 10 depending on whether k is + or -

now the quesiton is - would I have gotten that if ikaytas didn't give the answer... ;)
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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   09 Jun 2006, 08:30
But we don't know anything about a(1) since it is not defined...

a(1) could have been something bigger than 10...
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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink] New post   09 Jun 2006, 11:21
kook44 wrote:
it is b.

a[8] is the median. either a[1] thru a[7] are less than 10 or a[9] thru a[15] are less than 10 depending on whether k is + or -

now the quesiton is - would I have gotten that if ikaytas didn't give the answer... ;)


Good job kook44! :good



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Re: Consider the series : a(1), a(2), ....a(15). a(n) = a(n-1) + [#permalink]
09 Jun 2006, 11:21
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