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For any positive integer n, the sum of the first n positive

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For any positive integer n, the sum of the first n positive [#permalink] New post 01 Nov 2004, 07:12
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For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150
Senior Manager
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 [#permalink] New post 01 Nov 2004, 10:37
B.
between 99 and 301 there are 101 even Integers.
The middle one (Average) is 200.

200*101 = 20,200.
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 [#permalink] New post 01 Nov 2004, 10:45
101/2 [2*100 + (100) * 2] = 101 * 200 = 20200

dharmin
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 [#permalink] New post 01 Nov 2004, 10:52
Also got 20200 but my approach was less intuitive than Dookie's.
If you look at pattern of increase b/w first 5 positive integer, you get:
1-2-3-4-5
Now, the pattern of increase b/w first 5 even integer is:
2-4-6-8-10
If sum of first n positive integer is n(n+1)/2, then sum of first even n positive integer is 2n(n+1)/2 = n(n+1)
I looked for diff. b/w first 150 even integers (2-300 inclusive) and diff. b/w first 50(2-98 inclusive) even integers. I get 150(151) - 50(51) = 20200. Took me about 2 min.
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  [#permalink] 01 Nov 2004, 10:52
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