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by ashleyjan » Thu Oct 06, 2011 4:37 pm
Hi everyone,
I was wondering if someone could help me understand the logic behind the answer of this problem:
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 to 301?
A)10,100
B) 20200
C) 22650
D) 40200
E) 45150


The answer from the book says
=2(150(150+1)/2)-2(49(49+1)/2)
=150(151)-49(50)
=50(3(151)-49)
=50(453-49)
=20200

I just don't get how we got 150 and 49 from. I'll wait patiently for your response. Thank you in advance!
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by GmatMathPro » Thu Oct 06, 2011 4:44 pm
The sum of the even numbers from 99 to 301:

100+102+104.....+300=
2(50+51+52......+150)....note that inside the parentheses is the sum of 50 to 150. This is just the sum of the numbers from 1 to 150 minus the sum of the numbers from 1 to 49.

Once you find that, multiply everything by 2.

There are easier ways to solve it in my opinion, but that's where the 49 and 150 come from.
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by phoenix111 » Thu Oct 06, 2011 4:50 pm
Consider this :

Even numbr b/w 99 and 301
100..102...104......XXX...300

This is an arthimatic progression with difference(d) = 2

number of terms = (last term(l) - first term(a))/d + 1
= 200/2 + 1
= 101

Sum = n/2[ 2a + (n-1)d ]
= 101/2[ 200 + 100*2]
= 101/2 * 400 = 20200


Easier way : take average of number and multiply by number of terms :

(100 + 300)/2 * 101 = 20200
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by ashleyjan » Thu Oct 06, 2011 5:17 pm
Thank you guys for the speedy replies. It makes sense to me now. I really appreciate, thank you again.
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by smackmartine » Thu Oct 06, 2011 5:47 pm
We know that Sum = Avg * no. of terms
Also ,
(Sum of all even no.s from 99 to 301)
= (Sum of all even no.s from 1 to 301) - (Sum of all even no.s from 1 to 99)

= [(301+1)/2]*[(301-1)/2] - [(99+1)/2]*[(99-1)/2]
=22650 - 2450
=20200
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by GMATGuruNY » Thu Oct 06, 2011 7:23 pm
I posted a solution here:

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