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# If the sum of the first n positive integers is S , what is

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If the sum of the first n positive integers is S , what is [#permalink]

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29 Nov 2004, 07:25
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If the sum of the first n positive integers is S , what is the sum of the first n positive even integers, in terms of S ?

(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S

can someone pls explain ???
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29 Nov 2004, 07:48

First postive integers can be 1, 2, S = 3
First even positive integers will be 2,4, so sum will be S * 2
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29 Nov 2004, 08:11
Yep the ans is C

I picked numbers. I feel that is the easiest way.

Let n = 3

S will be 6 (1+2+3)

For the n first even integers:

2+4+6 = 12. Therefore S is now 2S. You can try other numbers and you will always get 2S as ur answer.

Hope this helps

Last edited by ninomoi on 29 Nov 2004, 10:32, edited 1 time in total.
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29 Nov 2004, 08:25
sum of first n positive integers is n(n+1)/2...

sum of first n even positive integers is n(n+1) which is 2S
how I arrived at n(n+1)

=2+4+6+......+2n
=2[1+2+3......+n]
=2[n(n+1)/2]
=n(n+1)
=2S

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29 Nov 2004, 12:22
I get C as well.

Sum of 1st n numbers S = n(n+1)/2

Sum of 1st n even numbers
= (n(4+(n-1)2))/2 (Using Arithmetic progression formula)
= n(n+1)
= 2S
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29 Nov 2004, 13:18
Mmm...why do we assume these n numbers are consecutive positive integers? What if they are in a random sequence like 1,2,5,8,13?
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29 Nov 2004, 13:38
christoph wrote:
If the sum of the first n positive integers is S , what is the sum of the first n positive even integers, in terms of S ?

The question asks for the first n +ve integers. n could 3 which means 1,2,3
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29 Nov 2004, 13:38
christoph wrote:
If the sum of the first n positive integers is S , what is the sum of the first n positive even integers, in terms of S ?

Junino

The question asks for the first n +ve integers. n could be = 3 which means 1,2,3 or n=5 which will be 1,2,3,4,5
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29 Nov 2004, 15:29
gayathri - What is the formula to calculate the SUM of n numbers in an Arithmetic Progression.
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29 Nov 2004, 16:36
gmat+obsessed wrote:
gayathri - What is the formula to calculate the SUM of n numbers in an Arithmetic Progression.

Sum of "n" numbers in an AP = (n/2)[2a+(n-1)d]
where
a = 1st number in the series
d = difference between consecutive terms in the series
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30 Nov 2004, 15:59
30 Nov 2004, 15:59
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