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In an infinite sequence of integers a1, a2, a3, , a1 = -30

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kevincan
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In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post   03 Dec 2007, 08:21
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In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four (B) Five (C) Six (D) Seven (E) Eight



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Re: In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post   14 Jan 2008, 22:43
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minimum k: -30+7*(k-1)=47 ==> k=77/7+1=12
maximum k: -30+5*(k-1)=47 ==> k=77/5+1=16.4 ==> k=16

Therefore, k e {12,13,14,15,16} ==> N=5
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Re: In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post   15 Jan 2008, 01:14
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kazakhb wrote:
why 7 and 5?


In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

the least step is +5 (because an-1 + 4 <an and an is integer)
the largest step is +7 (because an < an-1 +8 and an is integer)
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Re: In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post   15 Jan 2008, 10:08
a1=-30
using given equation an-1 + 4 <an < an-1 +8 we can find out range of values for a2
a1+4 < a2 < a1+8
-30+4 < a2 < -30+8
-26 <a2 < -22
this means a2 can have 5 values between -22 and -26. That means there can be 5 possible values of K....(using ak=47)
Answer is B
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Re: In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post   15 Jan 2008, 11:10
akhi wrote:
a1=-30
using given equation an-1 + 4 <an < an-1 +8 we can find out range of values for a2
a1+4 < a2 < a1+8
-30+4 < a2 < -30+8
-26 <a2 < -22
this means a2 can have 5 values between -22 and -26. That means there can be 5 possible values of K....(using ak=47)
Answer is B


excuse me, I believe a2 can have 3 values: we know that it must be greater than -26 and less than -22. it means that it could be -25, -24, -23. -26 and -22 are not included.


this is my opinion:

ak - a(k-1) could be a value included between 7 and 5
let's go with the difference of 7. we know that ak - a1=77. since the difference at limit is 7, we must divide 77 by 7 in order to find k, thus k=11

same reasoning for 5, thus we have 15 as top limit

values can be 11,12,13,14,15.



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Re: In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink]
15 Jan 2008, 11:10
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