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sum of even numbers

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by g.shankaran » Sun Jun 12, 2011 2:30 am
the sum of the first 50 +ve even integers is 2550. which is the sum of the even integers from 102 to 200 inclusive?

1. 5100
2. 7550
3. 10100
4. 15500
5. 20100
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by champ0007 » Sun Jun 12, 2011 2:42 am
no. of integers 50.

Sum of an AP = 50/2 * ( 102 + 200) = 25 * 302 = 7550
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by Calvin123 » Sun Jun 12, 2011 2:43 am
g.shankaran wrote:the sum of the first 50 +ve even integers is 2550. which is the sum of the even integers from 102 to 200 inclusive?

1. 5100
2. 7550
3. 10100
4. 15500
5. 20100
IMO 2.

This is problem of AP.
First term :102
Last term: 200
total terms: last-first/common difference +1 = 200-102/2 +1 = 50

Now sum of n terms of ap= n/2(first term + last term)
50/2(200+102)=> 25*302 => 7550
Hence 2. 7550
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by vinayreguri » Sun Jun 12, 2011 2:49 am
Sum of +ve even integers = n(n+1)=100(101)=10100 - (2550)sum of first 50 integers =7550
OR
((102+200)/2)*50=7550
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by GMATGuruNY » Sun Jun 12, 2011 3:36 am
g.shankaran wrote:the sum of the first 50 +ve even integers is 2550. which is the sum of the even integers from 102 to 200 inclusive?

1. 5100
2. 7550
3. 10100
4. 15500
5. 20100
To determine the sum of evenly spaced integers, use the following formula:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

With even integers, the distance between each successive pair is 2.
Thus, given the even integers from 102 to 200, inclusive:

Number of integers = (200-102)/2 + 1 = 50.
Average of biggest and smallest = (200+102)/2 = 151.
Sum = number * average = 50*151 = 7550.

The correct answer is B.
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by cans » Tue Jun 14, 2011 3:49 am
the sum of the first 50 +ve even integers is 2550. which is the sum of the even integers from 102 to 200 inclusive?
7550
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