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Arithmetic Estimation (sum of consecutive intergers)

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Arithmetic Estimation (sum of consecutive intergers)

by matteomasciotti » Sat Dec 22, 2012 9:48 am
Hi everyone,
I found this "simple" problem in the OG13 pretty annoying; it states the following;

M is the sum of the reciprocals of the consecutive integers from 201 to 300 inclusive. Which of the following is true?
A) 1/3 <M 1/2
B)1/5<M<1/3
C)1/7 <M< 1/5
D) 1/9 < M < 1/7
E) 1/12 <M< 1/9

So I decided to solve the problem this way:
1) find the midpoint in the sequence so (1/201 +1/300)/2
2) find the number of numbers in the sequence (300-201) +1
3) multiply the midpoint of the sequence times the number of numbers...and I get 0.83, which is not included in the options above.
I Looked at the solution at the end of the book and found a much simpler way to compute this whole summation...but...why is my techique wrong in this case?
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by Brent@GMATPrepNow » Sat Dec 22, 2012 11:52 am
matteomasciotti wrote:Hi everyone,
I found this "simple" problem in the OG13 pretty annoying; it states the following;

M is the sum of the reciprocals of the consecutive integers from 201 to 300 inclusive. Which of the following is true?
A) 1/3 <M 1/2
B)1/5<M<1/3
C)1/7 <M< 1/5
D) 1/9 < M < 1/7
E) 1/12 <M< 1/9
We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2 = A

Cheers,
Brent
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by matteomasciotti » Sat Dec 22, 2012 12:01 pm
Thanks Brent!
This is also the solution provided by the Official guide...But I was wondering why my method doesn't work in this case.
The number of numbers is still 100 in the sequence , even if the numbers are reciprocals of integers and hence fractions. So...?
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by Brent@GMATPrepNow » Sat Dec 22, 2012 12:10 pm
matteomasciotti wrote: So I decided to solve the problem this way:
1) find the midpoint in the sequence so (1/201 +1/300)/2
2) find the number of numbers in the sequence (300-201) +1
3) multiply the midpoint of the sequence times the number of numbers...and I get 0.83, which is not included in the options above.
I Looked at the solution at the end of the book and found a much simpler way to compute this whole summation...but...why is my techique wrong in this case?
The problem here is that you're treating the sequence 1/201, 1/202, 1/203, . . . 1/299, 1/300 as one that decreases at constant rate. In actuality, the difference between successive varies from term to term.

Granted, the sequence 201, 202, 203, . . . 300 increases at a constant rate (the difference between successive is always 1). So, one way to find the sum of these values is to find the midpoint (middle term) and then multiply this number by 100.

However, this strategy doesn't work if the sequence does not change at a constant rate.

Consider this example: {1, 1/10, 1/100}
Using your method, the "midpoint" is [1 + 1/100]/2 = 0.505
So, the sum = (3)(0.505) = 1.515

However, the real sum here is 1.11

I hope that helps.

Cheers,
Brent
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by matteomasciotti » Sat Dec 22, 2012 12:27 pm
Thanks Brent,
Now I got it
sorry about the "midpoint" :)
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by ritind » Mon Dec 24, 2012 11:05 pm
There's one more way
we know the terms are in AP
sum of terms = [no of terms(first term + last term)]/2
first term = 1/201
last term = 1/300
no of terms = 100
Put the values in formula
sum of terms = 0.41
option A 0.33<0.41<0.5
OA is A
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by puneetkhurana2000 » Mon Dec 24, 2012 11:12 pm
There's one more way
we know the terms are in AP
sum of terms = [no of terms(first term + last term)]/2
first term = 1/201
last term = 1/300
no of terms = 100
Put the values in formula
sum of terms = 0.41
option A 0.33<0.41<0.5
OA is A
The terms are not in AP, the question is about reciprocals. 1/202 - 1/201 is not equal to 1/203 - 1/202.

Hope this makes sense!!!
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