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Sum of all even integers...

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Sum of all even integers...

by money9111 » Mon Feb 01, 2010 8:28 pm
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. 10,000
b. 20,200
c. 22,650
d. 40,200
e. 45,150

again.. no idea where to start this one... definitely going to have to go through the beginning of the number properties books.
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by money9111 » Mon Feb 01, 2010 8:39 pm
woops OA B
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by papgust » Mon Feb 01, 2010 9:35 pm
Even integers range from 100 to 300. first, count the number of integers

(300-100)/2 + 1 = 101

Now find the average. Average = (First + Last)/2 = (100+300)/2 = 200

Sum of all even integers b/w 99 and 301 = 101 * 200 = 20,200
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by hai1 » Tue Feb 02, 2010 4:34 am
papgust, do you mean the number of even integers between 100 and 300 are 101?

To find this number, do you always have to use

For example between 1 & 20, the number of even integers would be (20-2)/2 +1=10
I think this also works for odd integers= (19-1)/2+1=10

Average for even= (20+2)/2= 11
Average for odd= (19+1)/2= 10

Sum of even's= 11*10=110
Sum of odd's=10*10=100

Is there a similar formula for product of odd or even integers? Please respond.
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by ajith » Tue Feb 02, 2010 7:16 am
money9111 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. 10,000
b. 20,200
c. 22,650
d. 40,200
e. 45,150

again.. no idea where to start this one... definitely going to have to go through the beginning of the number properties books.
Sum of first 'n' positive even integers = n(n+1)

sum of first 49 positive even integers = 49*50

sum of first 150 positive even integers = 150*151

sum of even integers between 99 and 301 = (150 * 151) - (49 * 50) = 20 200
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