sum of even integers

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sachindia: Master | Next Rank: 500 Posts; Posts: 258; Joined: Fri Jun 22, 2012 4:39 am; Location: Bengaluru, India; Thanked: 6 times; Followed by:3 members; GMAT Score:640

sum of even integers

by sachindia » Fri Jul 20, 2012 11:01 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 ?

Regards,
Sach

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Source: Beat The GMAT — Problem Solving | Fri Jul 20, 2012 11:01 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Fri Jul 20, 2012 11:43 pm

sachindia 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 ?

There is a much better way to solve this without using the given formula.

For any set of uniformly spaced integers, the average of the integers = (Largest integer + Smallest integer)/2

In this case, all the even integers between 99 and 300, i.e. from 100 to 300 are uniformly spaced. Hence, average of them = (100 + 300)/2 = 200
Now, sum of all these integers = (Average)*(Number of integers in the range) = 200*[(300 - 100)/2 + 1] = 200*101 = 20,200

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GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Fri Jul 20, 2012 11:51 pm

If anyone is interested in solving the problem using the formula...

Required sum = (100 + 102 + 104 + ... + 300) = 2(50 + 51 + ... + 150)

Now, (1 + 2 + ... + 150) = 150*151/2
And, (1 + 2 + ... + 49) = 49*50/2

Hence, (50 + 51 + ... + 150) = (150*151/2) - (49*50/2) = (50/2)*(3*151 - 49) = 50*404/2

Hence, required sum = 2*50*404/2 = 50*404 = 20,200