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Integers..

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by Elena89 » Mon Dec 12, 2011 7:00 am
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

Source: OG-12

[spoiler]OA: (B) [/spoiler]
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by GMATGuruNY » Mon Dec 12, 2011 7:21 am
I posted a solution here:

https://www.beatthegmat.com/for-any-posi ... 59401.html
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by neelgandham » Mon Dec 12, 2011 7:26 am
Hi Elena,

The problem can be solved using different methods. Here are a couple of them.
Method 1)
Consecutive even integers are in Arithmetic Progression, where the common difference of the series is 2, First term 100 and Last term 300.

nth term in an Arithmetic Progression = First term +(n-1)*(Common difference)
300 = 100 + (n-1)*2 => n = 101
Sum of n terms = (n/2)*(first term + last term) = (101/2)*(100+300) = 20200
Method 2)
sum of all the even integers between 99 and 301 = 100+102+104+....300
= 2*(50+51+52+...150)
= 2*((1+2+3+4+...150)-(1+2+3+.....49))
= 2*((0.5*150*151)-(0.5*49*50))
= ((150*151)-(49*50))
= (50*((151*3)-49))
= 50*404
= 20200
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by Elena89 » Mon Dec 12, 2011 7:30 am
Thanks Mitch.. Reading your post made me realize that I was using the right formula with just a tiny mistake! :wink:
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by ankush123251 » Thu Dec 15, 2011 2:04 am
Use the AP formula directly for sum with a1 = 100,an = 300 and d =2.
With this info find n and the Sn = n/2 * (a1 + an).
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