50/2( 2*102+ 49*2)
25( 204 + 98)
25( 302)
7550
Formulae used:
n/2( 2*a + (n-1)d)
n = no of terms of the AP
d = common difference
a = first term
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GMATPrep Q4
Source: Beat The GMAT — Problem Solving |
-
ronniecoleman
- Legendary Member
- Posts: 546
- Location: New Delhi , India
- Thanked: 13 times
Average of the set of evenly spaced consecutive integers = (1st no + last num)/2
in this case it will be (102+200)/2 = 151
# of even integers between 102 and 200 = 50
so the total = 151*50
=7550
in this case it will be (102+200)/2 = 151
# of even integers between 102 and 200 = 50
so the total = 151*50
=7550
I have no idea about all these formulas that are being used, but I just used common sense. We are told that the sum of the first 50 even positive integers {2, 4, 6... 100} is 2,550. To find the sum of the next 50 (which is what the question is asking), add 100 to each number in the original set: {2+100, 4+100, 6+100... 100+100}.
This equals 2,550 + 50(100) = 7,550, B
This equals 2,550 + 50(100) = 7,550, B
-
ronniecoleman
- Legendary Member
- Posts: 546
- Location: New Delhi , India
- Thanked: 13 times
common sense is no longer so common!!brb588 wrote:I have no idea about all these formulas that are being used, but I just used common sense. We are told that the sum of the first 50 even positive integers {2, 4, 6... 100} is 2,550. To find the sum of the next 50 (which is what the question is asking), add 100 to each number in the original set: {2+100, 4+100, 6+100... 100+100}.
This equals 2,550 + 50(100) = 7,550, B
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Looking at the problem, this is how the gmat test takers wanted u approach this problem. As u can see this might take longer to solve than the formulae based approaches Ronnie and amitabh have suggested.brb588 wrote:I have no idea about all these formulas that are being used, but I just used common sense. We are told that the sum of the first 50 even positive integers {2, 4, 6... 100} is 2,550. To find the sum of the next 50 (which is what the question is asking), add 100 to each number in the original set: {2+100, 4+100, 6+100... 100+100}.
This equals 2,550 + 50(100) = 7,550, B
Either way, adding these formulae to the arsenal will atleast provide a second path to the solution.
