even integers

Well, you could also do it this way:

The first 50 positive even integers are 2, 4, 6,..., 100.
The even integers from 102 to 200 will be 102, 104, 106,..., 200.

Just by looking at the list you can see the symmetry: the terms in the second list can be obtained by adding 100 to the terms in the first list. This means that the sum you're looking for is actually the first sum (2550) + 50 times 100 (since you add 100 to every term and you have 50 terms) = 2550 + 50*100 = 2550 + 5000 = 7550.

Of course, the problem can be solved by using your very own method. On test day, just pick whichever method comes to your mind and don't worry about anything else!

You have used a shortcut..

but if u calculate sum pf +ive even integers upto 200 inclusive.. it would be 200(200+1)/2*2 = 10100

To get sum of even integers between 102 & 200 inclusive, you need to subtract sum of even integers upto 100 (i.e. first 50 even integers) i.e. 2550

what you get is 10100 - 2550 = 7550

KICKGMATASS123 wrote:Just one question about this question itself....


The sum of the first 50 positive even integrs is 2550. What is the sum of the even integrs from 102 to 200 inclusive ??


5100
7550
10100
15500
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Why is the information of first positive 50 even intergers needed for this question..

I can just do this by Sum = Avg/ * n terms
Sum = 102+200/2 * 50 and I get 7550..

Is this information of the sum of 50 positive even integers extraneous or are we using it somewhere in the problem and I'm missing the point??

Thanks for your help..

Sum of first n positive even integers is n(n+1)
First 100 even is 100(101)=10100. So 10100-2550=7550