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

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The sum of the first 50 positive even integers is 2550. What [#permalink]  12 Jun 2008, 03:54
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The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10100
15500
20100

I know that there are 200 -102 =98 = 49 even numbers.

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Re: Sum of even integers [#permalink]  12 Jun 2008, 05:55
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10100
15500
20100

I know that there are 200 -102 =98 = 49 even numbers.

actually there are 50 even numbers (200-102)/2+1 /// think about counting 1, 2, 3 ... 10 (10 -1 = 9 but there are 10 numbers)
i would say the sum is equal to 2*(sum of all numbers between 51 and 100) = 2*(100+51)*50/2=151*50=7550 -> B
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Re: Sum of even integers [#permalink]  13 Jun 2008, 03:40
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10100
15500
20100

I know that there are 200 -102 =98 = 49 even numbers.

this will be an AP series question
102,104,106....200

a = 102 (the first term)
d = 2 (the common difference)
n = 50 (total even numbers for which the sum will be calculated)

now the formula for sum of an AP series is

S = N/2(2a + (n-1)d)

= 50/2 (2*102 + (50-1)2)

=25(204+98)
=25*302
=7,550

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Re: Sum of even integers [#permalink]  13 Jun 2008, 03:51
A more simple way to solve that without knowing any formula is to use what they say in the question :

sum of [even integers between 102 and 200 (inclusive)] = sum of [100 + even integers between 2 and 100 (inclusive)] = 50*100 + sum of the first 50 positive even integers = 5000 + 2550 = 7550
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Re: Sum of even integers [#permalink]  02 Jul 2008, 03:22
we can use elimination.

1. upper cap 50*200 =10000 therefore option C/D/E are out. Now choose between A and E

2. There are 50 even numbers between 102-200 including 102 therefore the sume should be greater than 50*100=5000.

3. The sum has to be more than 5100 therefore only choice left is 7550
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Re: Sum of even integers [#permalink]  02 Jul 2008, 03:54
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

5100
7550
10100
15500
20100

I know that there are 200 -102 =98 = 49 even numbers.

I do not understand why they give us "The sum of the first 50 positive even integers is 2550".

We can find "the sum of the even integers from 102 to 200" without any additional information.
200-102 = 98 ==> Their are 50 even-numbers in the set.
S=50*(102*2+49*2)/2 = 7550.
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Re: Sum of even integers [#permalink]  02 Jul 2008, 04:09
lexis wrote:
I do not understand why they give us "The sum of the first 50 positive even integers is 2550".

We can find "the sum of the even integers from 102 to 200" without any additional information.
200-102 = 98 ==> Their are 50 even-numbers in the set.
S=50*(102*2+49*2)/2 = 7550.

Yes we can. But even without knowing the formula you used we could have solved the question with the piece of information they give us.

As I wrote above :

$$\sum_{n=102}^{200} n_{even} = \sum_{k=51}^{100} 2k = \sum_{k=1}^{50} (100+2k) = \sum_{k=1}^{50} 100 + \sum_{k=1}^{50} 2k = 50*100 + 2550 = 7550$$

(without calculating anything )
Re: Sum of even integers   [#permalink] 02 Jul 2008, 04:09
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