jakolik wrote:

crack700 wrote:

Can you post the exact problem here

Crack 700, below is question 157 from

OG12For 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

Easy...

question asks what is...

100+102+104.......+300

We can solve it in 2 ways...

First MethodAverage of 100 and 300 is 200

Total even number between 100 and 300 (inclusive) = [(300-100)/2] + 1 = 101

therefore 101*200 = 20200 (ans B)

Second Method100+102+104.......+300

=2(50+51+52+....+150)

Sum of (50+51+52+....+150) equals

Sum of (1+2+3.....+150) - (1+2+3+...50)

= 11325-1275

= 10050

Now we have the answer as 2*10050 = 20200

Hope this helps.....

_________________

CONSIDER AWARDING KUDOS IF MY POST HELPS !!!