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For positive integers n, the integer part of the nth term of sequence

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For positive integers n, the integer part of the nth term of sequence  [#permalink]

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New post 24 Apr 2015, 03:17
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A
B
C
D
E

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Question Stats:

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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.

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Re: For positive integers n, the integer part of the nth term of sequence  [#permalink]

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New post 24 Apr 2015, 03:36
1
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  [#permalink]

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New post 24 Apr 2015, 10:29
1
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  [#permalink]

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New post 24 Apr 2015, 11:03
1
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  [#permalink]

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New post 24 Apr 2015, 17:16
1
1
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  [#permalink]

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New post 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  [#permalink]

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Re: For positive integers n, the integer part of the nth term of sequence   [#permalink] 26 Aug 2018, 09:02
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