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OG Problem Solving

by lola27 » Sat Jan 28, 2012 3:12 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 even integers between 99 and 301?

a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150

can anyone share an easy way of doing this.. i dont really understand the explanation in the book. Thanks!
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by neelgandham » Sat Jan 28, 2012 3:31 pm
Method 1:
The question Find the sum of the EVEN integers from 99 to 301 is same as the question Find the sum of the EVEN integers from 100 to 300 (both Inclusive).
Find the average of the set (First Term + Last Term)/2 (100+300)/2=200
Find the number of terms (Last Term - First Term)/2 + 1 = (300-100)/2 +1= 101
Multiple the average of the number to the number of terms to get answer : 200*101=20,200
Method 2:
The series 100,102,.....300 are in AP. Let the number of terms be n, then the
nth term = first term + ((n-1)* common difference of successive members) = 300 = 100 + (n-1)*2 => n-1 = 100 => n = 101
Sum of all the terms in an AP is equal to 0.5*n*(first term + Last term) = 0.5*101*400 = 20200
Method 3:
Let s = 100+102+104+....300 = 2*(50+51+....150) = 2*(Sum of first 150 positive integers - Sum of first 49 positive integers) = 2*((0.5*150*151)-(0.5*49*50)) = 20200

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by GMATGuruNY » Sat Jan 28, 2012 4:26 pm
I posted a solution here:

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