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In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink]
 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, a n-1 + 4 <a n < a n-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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Bminimum 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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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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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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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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