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Senior Manager
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New post 26 May 2006, 14:08
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The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

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

Please explain your answers?
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New post 26 May 2006, 14:25
102 + 104 + ... + 200 = 100 * 50 + 2 * ( 1 + 2 + ... + 50 ) = 5000 + 2 * 2550 = 10100
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New post 26 May 2006, 22:29
Here is my piece.
From 102 to 200 inclusive, there are 99 integers, 50 even and 49 odd. The sum of the first even and last even is
102+200=302
104+198=302
106+196=302 and so on you have got 25 pairs with sum 302 or 7550
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New post 27 May 2006, 00:44
BG wrote:
Here is my piece.
From 102 to 200 inclusive, there are 99 integers, 50 even and 49 odd. The sum of the first even and last even is
102+200=302
104+198=302
106+196=302 and so on you have got 25 pairs with sum 302 or 7550


You are right, I got the question wrong since 2550 is not the sum of
first 50 integers, but that of first 50 even integers ( 2+4+...+100 )

Well done!
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New post 27 May 2006, 03:39
I got B. Just add 100 to each of the previous 50 even integers. Since there are 50 terms, you add 5000 to the previous total, which makes B the logical answer.
  [#permalink] 27 May 2006, 03:39
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