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# The sum of the first 50 positive even integers is 2550. What

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Manager
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The sum of the first 50 positive even integers is 2550. What [#permalink]

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15 Apr 2006, 14:37
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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 - 200, inclusive?

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Manager
Joined: 09 Feb 2006
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15 Apr 2006, 15:12
You don't need to really know the first part of the question, however there may be a trick with it. I know my own way, which is fine by me.

First, find the average of the integers. 200+102 = 302/2 = 151

Second, find the number of even integers: 200-102 = 98/2 (to get the evens only) = 49 + 1 (because it is inclusive) = 50.

Then multiply the average times the number of even integers, to get: 7550.

SORRY: EDITED FOR MY POOR GRAMMAR

Last edited by jcgoodchild on 15 Apr 2006, 16:29, edited 1 time in total.

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Manager
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18 Apr 2006, 00:31
There is a reason why the first piece of information was given. According to the question -

2 + 4 + 6 + ....+ 100 = 2,550. Now find the following -
102 + 104 + 106 + ....+ 200 = ?

You can easily see that each term in the second expression is equal to 100 plus the term in the first expression. So the second expression can be written as -

(100 +2) + (100 + 4) + ......+ (100 + 100) = 50*100 + First Expression = 5,000 + 2,550 = 7,550
_________________

Thanks,
Zooroopa

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Manager
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18 Apr 2006, 00:38
Nice touch, Zooroopa! So we now have two good and easy ways to solve Q's like this one...

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VP
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18 Apr 2006, 00:47
You do not need the sum of the first 50 even integer.. Even otherwise u can find the value

Its an arithmetic progression with a (first value) =102
d (difference) =2
l (Last value) =200

The number of terms: n = (200 -102)/2 + 1= 50

Sum of an AP = n/2*(a+l)

= 50/2*(102+200)
= 7550

Kudos [?]: 29 [0], given: 0

18 Apr 2006, 00:47
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