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Working with sequences

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Working with sequences [#permalink] New post 12 Jul 2011, 04:00
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (02:07) correct 40% (02:13) wrong based on 15 sessions
Hi everybody,

I'm having a hard time going through sequences for some reason, so your explanations will be greatly appreciated. It goes like this:

The sequece S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/n+1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

1. k>10
2. k<19


Thanks.

-R
[Reveal] Spoiler: OA
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Re: Working with sequences [#permalink] New post 12 Jul 2011, 11:51
First see the sum of first 3 terms: (1 - 1/2) + (1/2 - 1/3 ) + (1/3 - 1/4) = 3/4
similarly first sum of first 2 terms: 2/3
therefore it will be of the form, sum of first n terms = n/n+1
since this is a proper fraction, adding one to Numerator and denominator increases the overall value.
consider stmt 1: k > 10; the sum of first 10 terms is = 10/11 which is (9+1)/(10+1) that means greater than 9/10.

but in stmt 2: sum of fist 9 (or fewer terms) is less than 9/10 and sum of terms more than 9 is greater than 9/10.

therefore answer will be A
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Re: Working with sequences [#permalink] New post 14 Jul 2011, 05:06
I could not do this in 2mins. picked a wrong strategy.. damn :twisted:

Sn = 1/n - 1/(n+1)

Sum of first k terms = S1 + S2 ... Sk = 1/1 -1/2 + 1/2 - 1/3 ... + 1/(k-1) - 1/k + 1/k - 1(k+1) = 1 - 1/(k+1) = k/(k+1)

St#1 : k>10 , assume k=11, then Sum = 11/12, which is > 9/10, (also subsequent sums for k = 12, 13, 14 etc. will be greater that 9/10). Suff. there possible options AD

St #2: k<19, assume k=2, then Sum = 2/3, which is < 9/10 and we already saw that for k=11, sum > 9/10 so not suff. eliminate D => A.
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Re: Working with sequences   [#permalink] 14 Jul 2011, 05:06
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