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For positive integers n, the integer part of the nth term of sequence
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24 Apr 2015, 03:17
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53% (02:30) correct 47% (02:37) wrong based on 162 sessions
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Re: For positive integers n, the integer part of the nth term of sequence
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24 Apr 2015, 03:36
A1=1.23456 A2=2.468 A3=3.91215 A4=4.81216 A5=5.1015 A6=6.1218 A7=7.1421
have sum of 7 consecutives, so (1+7/2)*7=28, and sum of decimals equal to 26
so 28+2.6=30.6
C



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Re: For positive integers n, the integer part of the nth term of sequence
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24 Apr 2015, 10:29
1.2345 2.4681 3.6912 4.8121 5.1015 6.1218 7.1421 total = 30.5713 hence between 30 and 31 Answer C
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Re: For positive integers n, the integer part of the nth term of sequence
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24 Apr 2015, 11:03
Bunuel wrote: For positive integers n, the integer part of the nth term of sequence A equals n, while the infinite decimal part of the nth term is constructed in order out of the consecutive positive multiples of n, beginning with 2n. For instance, A_1 = 1.2345678…, while A_2 = 2.4681012… The sum of the first seven terms of sequence A is between:
A. 28 and 29 B. 29 and 30 C. 30 and 31 D. 31 and 32 E. 32 and 33
Kudos for a correct solution. For simplicity of addition, i rounded off the digits to 1 decimal place => 1.2 + 2.4 + 3.6 + 4.8 + 5.1 + 6.1 + 7.1 = 30.3 Hence, answer is C



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Re: For positive integers n, the integer part of the nth term of sequence
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24 Apr 2015, 17:16
1.2345=1.2 2.46810=2.5 3.6912=3.7 4.812=4.8 5.1015=5.1 6.1218=6.1 7.1421=7.1
sum is 30.5
Answer: C



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Re: For positive integers n, the integer part of the nth term of sequence
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27 Apr 2015, 01:52
Bunuel wrote: For positive integers n, the integer part of the nth term of sequence A equals n, while the infinite decimal part of the nth term is constructed in order out of the consecutive positive multiples of n, beginning with 2n. For instance, A_1 = 1.2345678…, while A_2 = 2.4681012… The sum of the first seven terms of sequence A is between:
A. 28 and 29 B. 29 and 30 C. 30 and 31 D. 31 and 32 E. 32 and 33
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:First, construct the first seven terms, though only out to a few decimal places, following the given pattern. A_1 = 1.23… A_2 = 2.46… A_3 = 3.69… A_4 = 4.812… A_5 = 5.10… A_6 = 6.12… A_7 = 7.14… Now, to add up the first seven terms, you should be strategic about how many decimal places to keep. You can drop the hundredths place and get a good approximation with the tenths place—and if you find the sum too close to a boundary between choices, then you can refine your answer if necessary. 1.2 + 2.4 + 3.6 + 4.8 + 5.1 + 6.1 + 7.1 = 30.3 Including more decimal places would only add a couple of tenths to the sum—not enough to tip the sum over 31. The correct answer is C.
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Re: For positive integers n, the integer part of the nth term of sequence
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