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

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Director
Joined: 19 Mar 2007
Posts: 522
For any positive integer n, the sum of the first n positive [#permalink]

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11 Jun 2007, 08:16
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For any positive integer n, the sum of the first n positive integers equals 2. What is the sum of all integers between 8 and 30?

Last edited by nick_sun on 28 Jun 2008, 02:31, edited 1 time in total.
Director
Joined: 26 Feb 2006
Posts: 899
Re: GSET 10 - integers between 99 and 301 (PS) [#permalink]

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11 Jun 2007, 09:02
nick_sun wrote:
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

n = 101

= (301 - 99)/2 (301 + 99)/2
= 101 (400)/2
= 101 (200)
= 20,200
VP
Joined: 08 Jun 2005
Posts: 1145

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11 Jun 2007, 10:41
[(150*151/2)-(49*50/2)]*2 = 20,200

We need the sum of positive numbers between 99 and 301. we can find the sum of numbers from 50 to 150 and multiply by 2.

since:

50*2 = 100 (the first positive number)
51*2 = 102 (the second positive number)

and so on

150*2 = 300 (the last positive number)

to find the sum of numbers from 50 to 150 we will use the given formula:

150*151/2 = the sum of all number from 0 to 150. subtract 49*50/2 or sum of numbers smaller then 50.

now multiply by 2 to get the sum of positive numbers from 100 up to 300.

Intern
Joined: 02 Jun 2007
Posts: 27

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11 Jun 2007, 11:14
Here's a nifty formula you can use for this:

Sum of sequence of numbers = (first + last) * [(number of numbers)/2]

In this case it will be: (100 + 300) / [ (300-100)/2 + 1) / 2]

=> 400 / (101/2)
=> 400*101/2
=> 200*101
=> 20200
11 Jun 2007, 11:14
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