Hi Phelps (Michael?):

This is a great question - do you know where it's from? I ask because the first sentence seems out of place although it's a great tip. I haven't seen the GMAT give away a tip like that as part of the question (it almost reads as though it's setting up a sequence problem), and in this case using the tip is probably a lot more cumbersome than anything else.

Sum of consecutive even/odd/overall integers problems have come up fairly commonly in GMAT resources, and it's actually a statistics rule that can help you the most: In evenly-spaced sets, the median of the set will equal the mean. Knowing that means that if you can find the middle number, then you know the average of the set, and can then just multiply the average by the number of terms to get the total:

Average = sum of terms / number of terms

So, using algebra:

Average * number of terms = sum of terms

Here, for even integers, we'll start with 100 and end with 300 (and it's an inclusive set, since both 100 and 300 are within the parameters of 99 and 301), so the middle number is 200.

Now we need to find the number of terms. Because we only care about all of the even numbers, but not the odds, we only care about half of the numbers (every other number). So we can take the range (300-100 = 200) and divide by 2 to get rid of the ones we don't want (the odd ones), giving us a "usable" range of 100 values*.

Because it's an inclusive set, we need to add one to the number of values. It's a rule that you could memorize (inclusive --> add 1; exclusive --> subtract 1), or you could just prove it to yourself with the set 1, 2, 3 (the range is 2, but we know that there are 3 values if we include them all and only 1 if we don't include the endpoints, so you can prove that you add for inclusive and subtract for exclusive).

Adding 1 to the "usable" range of 100 gives us 101 values, and if multiply that by the average of the values, 200, we get:

100*200 + 1 * 200
20000 + 200
20,200 = B


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*Try this on a number line and it may make more sense. If you only care about, say, the multiples of 3 from 3 to 15, you'd list the numbers as:

3 4 5 6 7 8 9 10 11 12 13 14 15

If you divide by 3, because every third number gives you what you want, you'd have:

1 4/3 5/3 2 7/3 8/3 3 10/3 11/3 4 13/3 14/3 5

The integers that are left are the numbers that are divisible by 3, so you have 5 values, and you can then apply the inclusive/exclusive rules there.

Sum of even integers<=n

If n is odd: (n-1)/2*(n+1)/2

Sum of even intetgers<=99 (98)/2*(100)/2=2450

Sum of even intetgers<=301 (300)/2*(302)/2=22650

Sum of even integers between 99 and 301=22650-2450

20200

Option B