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by winniethepooh » Fri Jul 08, 2011 4:28 pm
For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
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by Ashley@VeritasPrep » Fri Jul 08, 2011 4:40 pm
You want the sum of the even integers 100, 102, 104, ..., 296, 298, 300.

One method. Pair up these numbers working from the ends in, e.g. 100, 102, 104, ..., 296, 298, 300. Notice that each pair has the same sum, 400. Now determine how many pairs there are. Well, there are 201 integers from 100-300 inclusive, and 101 of them are even. So there are 101 numbers in the list, which means half that many -- 50.5 -- pairs. (For these purposes, the math works just fine in terms of thinking of the existence of half a pair, even though that's sort of a questionable entity in the world of (say) socks.) Each pair is worth 400 and there are 50.5 of them ---> 50.5*400 = 20,200.
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by Ashley@VeritasPrep » Fri Jul 08, 2011 4:44 pm
Another method: This is an evenly spaced set (i.e. all the intervals between entries are identical), so it's easy to find its average value: it will just be the middle entry, 200. The sum of all the entries will equal the number of entries times the average (think of the average formula: average = (sum of entries)/(number of entries) ---> therefore, average * number of entries = sum of entries). As mentioned in the previous method, there are 101 entries in this list, so 101 * 200 = 20,200.
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by Ian Stewart » Fri Jul 08, 2011 6:23 pm
winniethepooh wrote:For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
Again, since you want additional approaches (the ones Ashley suggests are perfectly good, and are most likely what I'd use), you can also estimate here easily enough. We're adding roughly half of the numbers from 100 to 300, so we're adding roughly 100 numbers with an average of 200, so the sum should be roughly 100*200 = 20,000. This estimate is only off because we left out one number, so B will be correct.

There are also many ways to rewrite the sum. We need to find:

100 + 102 + ... + 298 + 300 = 2(50 + 51 + ... + 149 + 150) = 2[(49 + 1) + (49 + 2) + (49 + 3) + ... + (49 + 100) + (49 + 101)]

Since we are adding the numbers from 1 through 101 in brackets, we have 101 forty-nines, so this is equal to

2[101*49 + 1 + 2 + 3 + ... + 100 + 101]

Now we can use the formula in the question with n=101:

= 2[101*49 + (101)(102)/2] = (2)(101)(49) + (101)(102) = (2)(101)[49 + 51] = 202*100 = 20,200

That's probably not the easiest approach, but you asked for "as many solutions as possible!"
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by GMATGuruNY » Sat Jul 09, 2011 2:50 am
Check here:

https://www.beatthegmat.com/sum-of-even- ... 73600.html

The solution I offered (not unlike Ashley's) describes how to determine the sum of any set of evenly spaced integers.
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by Buix0065 » Sat Jul 09, 2011 3:16 am
Ashley@VeritasPrep wrote:Another method: This is an evenly spaced set (i.e. all the intervals between entries are identical), so it's easy to find its average value: it will just be the middle entry, 200. The sum of all the entries will equal the number of entries times the average (think of the average formula: average = (sum of entries)/(number of entries) ---> therefore, average * number of entries = sum of entries). As mentioned in the previous method, there are 101 entries in this list, so 101 * 200 = 20,200.
Ashley,

This is how I would have thought to approach this problem, if the problem had asked for the sum of even integers between 99 and 300... but by throwing in that formula where the sum of the first positive integer equals n(n+1)/2. How do you interpret this equation? What does seeing this equation change in the approach vs. if they just asked for the sum?

Thank you!
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by rveeraga » Sat Jul 09, 2011 7:58 pm
Buix0065 wrote:
Ashley@VeritasPrep wrote:Another method: This is an evenly spaced set (i.e. all the intervals between entries are identical), so it's easy to find its average value: it will just be the middle entry, 200. The sum of all the entries will equal the number of entries times the average (think of the average formula: average = (sum of entries)/(number of entries) ---> therefore, average * number of entries = sum of entries). As mentioned in the previous method, there are 101 entries in this list, so 101 * 200 = 20,200.
Ashley,

This is how I would have thought to approach this problem, if the problem had asked for the sum of even integers between 99 and 300... but by throwing in that formula where the sum of the first positive integer equals n(n+1)/2. How do you interpret this equation? What does seeing this equation change in the approach vs. if they just asked for the sum?

Thank you!
The question to ask yourself is, what do I know about the equation? If the answer is none, simply ignore that and follow the approach that you already know to find the sum of integers.

As you may know, n(n+1)/2 is used to find the sum of a series of n integers starting from 1. So, that equation might help those, who do not know the shortcut, to convert the even number sequence between 99 and 301 to all numbers sequence from 1 to 301 and then subtract the numbers that are not part of the finding: a long approach and tedious process for the question in hand.

As indicated earlier by Ashley and Mitch, the approach-sum=number of terms*average of numbers-is obviously the quickest to find an answer for the question in hand.

I will specify a high end formula to find the sum of the terms of an arithmetic sequence
S = n/2[2*a+(n-1)d], in which
n = number of terms = 101
a = first term = 100
d = the difference between consecutive terms = 2
So, S = 101/2 [2*100 + 100*2] = 20,200
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