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Sequences

by sparkle6 » Wed Sep 21, 2011 7:51 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
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by GMATGuruNY » Wed Sep 21, 2011 7:56 am
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by gmatclubmember » Wed Sep 21, 2011 7:57 am
The no. of terms would be 101 [(300-100)/2]+1
Sum of n terms = 101/2[200+100*2]=20200.

B is the answer.

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by cavaliar » Wed Sep 21, 2011 8:04 am
Answer B

Sequence between 2*50 and 2*150 even numbers

so 2*( [(150*151)/2]- [(50*51)/2] + 100

=20200
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by Anurag@Gurome » Wed Sep 21, 2011 7:17 pm
sparkle6 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

We need the sum 100 + 102 + 104 +.....+300 = 100 + (100 + 2) + (100 + 4)....(100 + 200)
= 100*101 + 2*(1 + 2 +..+100) = 100*101 + 2*100*101/2 = 2*100*101 = 20,200
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