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Bunuel
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Bunuel
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

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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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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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Bunuel
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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Looks like an Ap but its not an AP

Approximate the values by rounding off and the n do summation
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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_1 = 1.2345678...
A_2 = 2.481012...
A_3 = 3.691215...


A_7 = 7.142842...

A_1 + A_2 + .... + A_7 = 1+2+3+...+7 + .2345 + .4810 + .6912 + .81632 + .1020 + .1224 + .1428 + x
where x<<1

A_1 + A_2 + .... + A_7 = 28 + 2.6 + y = 30.6 + y; y<<1

IMO C
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