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
[spoiler]OA=C[/spoiler]
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For positive integers n, the integer part of the nth term
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The digits long after the decimal point are not going to matter here, since they only make a minuscule contribution to the sum, and we only need a rough estimate. If we work out the first seven terms, truncating each to one decimal place, we have:
1.2
2.4
3.6
4.8
5.1
6.1
7.1
When we round each term down here by ignoring all of the digits after the tenths digit, we are reducing the value of each term by less than 0.1, so if we add the seven terms above, the estimate we get will be less than (7)(0.1) = 0.7 below the true sum. And if we add the seven terms above, we get 30.3, so the real sum is somewhere between 30.3 and 31.0, and C is the answer.
1.2
2.4
3.6
4.8
5.1
6.1
7.1
When we round each term down here by ignoring all of the digits after the tenths digit, we are reducing the value of each term by less than 0.1, so if we add the seven terms above, the estimate we get will be less than (7)(0.1) = 0.7 below the true sum. And if we add the seven terms above, we get 30.3, so the real sum is somewhere between 30.3 and 31.0, and C is the answer.
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