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The sum of the first 50 positive even integers is 2550. what

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by harsh.champ » Thu Feb 18, 2010 12:34 pm
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
Sum = 2 x [51 + 52 + .... + 100]
Let the sum be 2x and solve. {Here x = [+ 52 + .... + 100 }
Apply the formula(sum of n natural no.s)
Sum of 1st 100 no.s = 100(100+1)/2 = 5050
Sum of 1st 50 no.s = 50(51)/2 = 1275
Hence we get x = 5050 - 1275 = 3775
Hence,the answer will be [spoiler]2 x 3775 = 7550 (B)[/spoiler]
Last edited by harsh.champ on Thu Feb 18, 2010 12:44 pm, edited 1 time in total.
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by Osirus@VeritasPrep » Thu Feb 18, 2010 12:44 pm
Did you type the original question wrong? Should it read 100 to 200? Otherwise none of the answers are correct.

To find the sum, find the number of terms and multiply that by the avg of the first and last. To find the number of terms the formula is (first - last)/interval. in this case it would be 200-102/2 = 49. The average of 200 and 102 is 151. Therefore the answer would be 151 * 49 = 7399
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by ajith » Thu Feb 18, 2010 12:56 pm
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
100*50 +2550 = 7550
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by imane81 » Fri Feb 19, 2010 5:43 pm
ajith wrote:
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
100*50 +2550 = 7550

Osirus, i found your approach very interesting. I think you would have fit with the answer B if you have had the following formula to count the number of the n digits within a set which is equal i think to

last term of the set - the first term of the set + 1

Applying this to the set from 102 to 200 and knowing we take into account only even numbers we found : (200 - 102)/2 + 1 = 50
Then according to your formula, 50* (200+102)/2 = 7550
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by thephoenix » Fri Feb 19, 2010 7:32 pm
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
total terms=(200-102+1)/2=50

sum=50/2[102+200]=7550
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by shashank.ism » Sat Feb 20, 2010 1:34 am
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
you can solve it easily by AP methos
i. e. sum of numbers = n/2 (1st term + last term)
here from 102 to 200 we have total n= 50
so sum = 50/2 (102 + 200) = 25 * 302 =7550 Ans B
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by Strongt » Mon May 30, 2011 12:55 pm
thephoenix wrote:
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
total terms=(200-102+1)/2=50

sum=50/2[102+200]=7550
when i divide (200-102 + 1) /2 i get 49.5!
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by ajith » Mon May 30, 2011 12:59 pm
Strongt wrote:
thephoenix wrote:
imane81 wrote:Please help on this one

The sum of the first 50 positive even integers is 2550. what is the sum of the even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
total terms=(200-102+1)/2=50

sum=50/2[102+200]=7550
when i divide (200-102 + 1) /2 i get 49.5!
It should have been (200-102)/2 +1
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by shashank.ism » Thu Jun 30, 2011 3:00 am
Its quite clear that the number of terms is 50.
102, 104, ..., 200 . i.e. 2x(51, 52,......, 100)
Now what is the total number of terms from 51 to 100 .. Its 50.
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