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# Number systems

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04 Jan 2012, 17:58
1/100 + 1/101 + 1/102 + ..... 1/110?.

Can somebody let me know, how to solve these kind of problems?.
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04 Jan 2012, 19:08
There is really no quick and dirty way to do this that I know of. You can get a close approximation by taking the average value of the first and last terms and multiplying by the number of terms (in this case 11*1/105). If you need to be exact this can be done by adding all permutations of the denominators in the numerator and the new denominator will be the product of all the old denominators. Perhaps if you posted a few answer choices more specific help could be given.
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07 Jan 2012, 13:15
Yeah, this would be kind of a disaster to solve precisely, so you're not likely to see it on the GMAT. However, you might see something where each of the terms had a common factor that you could pull out, or something like that. Did you have a particular problem in mind?
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07 Jan 2012, 14:04
you can get the sum by multiplying the average of numbers by the number of elements

sum=average of numbers * number of elements

we have 11 elements 100 to 110===> (110-100+1=11)
if we have a consecutive odd number of elements then the average is the middle number which 105

so sum is the product of 1/105 time 11
so, 11/105
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07 Jan 2012, 14:15
manalq8, that approach only works for consecutive integers, because they are evenly spaced. Since the consecutive integers are in the denominator here, the terms are not evenly spaced (i.e. the difference between 1/100 and 1/101 is not the same as the difference between 1/101 and 1/102). As joeshmo pointed out, the averaging approach would not get us a precise answer for this problem.

To illustrate, let's look at a simpler set of fractions:

1/1 + 1/2 + 1/3 =

If we multiply the median by the number of terms, we'd say the total is 3/2. However, the actual total is 11/6, which is about 22% greater than the projected answer. Also, notice that in this case, the average of the first and last terms does not equal the median. (1/1 + 1/3)/2 = (4/3)/2=4/6 = 2/3 If we average all 3 terms, we get yet another result (11/18). You can try this with any set of fractions and you'll get similar results. The consecutive numebr tricks are great, but they only work if the actual terms in question are evenly spaced!
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13 Jan 2012, 16:38
How would you actually solve this problem?
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Re: Number systems   [#permalink] 13 Jan 2012, 16:38
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