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the sum of integers

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Intern
Intern
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Joined: 05 Feb 2010
Posts: 18
Followers: 0

Kudos [?]: 1 [0], given: 2

the sum of integers [#permalink] New post 22 Feb 2010, 11:01
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (03:28) correct 0% (00:00) wrong based on 3 sessions
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10,100
15500
20100
Please explain. Thanks
Manager
Manager
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Joined: 26 May 2005
Posts: 210
Followers: 2

Kudos [?]: 71 [0], given: 1

Re: the sum of integers [#permalink] New post 22 Feb 2010, 11:12
Ekin4112 wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10,100
15500
20100
Please explain. Thanks

sum of even integers from 102 to 200 .. we can do that in 2 ways .. using information from the questions or just using the sum of consequetive intergers formula
102 + 104 ... + 200 = 100 + 2 + 100 + 4 + ..... 100 + 100 = 5 * 100 + 2 + 4 + ... 100 = 5000+ 2550(from question)=7550
sum = n/2 [2a + (n-1) * d] = n/2 [first + last term] = 50/2 * [102 + 200] = 7550

B
Senior Manager
Senior Manager
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Joined: 01 Feb 2010
Posts: 268
Followers: 1

Kudos [?]: 35 [0], given: 2

Re: the sum of integers [#permalink] New post 23 Feb 2010, 21:04
Ekin4112 wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10,100
15500
20100
Please explain. Thanks


Given by problem stmt is sum of first 50 even integers is 2550.
Needed is sum of even integers from 102 to 200 in other words next 50 even integers.
Difference between 2(1st even integer) and 102(51st even integer) is 100
Difference between 4(2nd even integer) and 104(52nd even integer) is 100

same way if we add all the differences then it will be 100*50 = 5000
5000+2550 = 7550 hence B
Manager
Manager
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Joined: 09 Dec 2009
Posts: 122
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Kudos [?]: 25 [0], given: 19

Re: the sum of integers [#permalink] New post 24 Feb 2010, 06:48
Here's how I solved:

Evenly spaced set, so we can find the average value by adding first term + last term and dividing by 2.
so, (102+200)/2 = 151.

We find the number of terms by taking the Last term - First term and ADDING one. Now here is the small trick.
So, 200-102 = 98. But its even numbers only so we divide by 2 which = 49, then add one. So we have 50 terms that we multiply by the sets average value - 151.

So 151*50 = 7550.

I vote B.
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Re: the sum of integers   [#permalink] 24 Feb 2010, 06:48
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