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The sum of the even numbers between 1 and n is 79*80, where

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Manager
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The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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New post 28 Oct 2008, 05:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

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Manager
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Re: Please help # 1/4 [#permalink]

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New post 28 Oct 2008, 06:58
It's the right answer, however, could you please explain?

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Director
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New post 28 Oct 2008, 07:09
Natasha0123 wrote:
It's the right answer, however, could you please explain?


Arithmetic progression with d=2

sum to first n numbers of Arithmetic progression= n/2[2a+(n-1)d] = n/2 [ 4 + 2n -2 ]
= n*(n+1)
= 79 * 80

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Re: Please help # 1/4 [#permalink]

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New post 29 Oct 2008, 04:06
Where am I going wrong in my approach?

Sum = 2 + 4 + 6 + ....+ (n-1)
= 2[1 + 2 + 3 + .....+ (n-1)/2]
= 2*(n-1)/2* (n+1)/2*1/2 = (n-1)(n+1)/4 = 79*80
or, (n-1)(n+1) = 79*80*4 and this will not give n = 79.

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Re: Please help # 1/4 [#permalink]

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New post 29 Oct 2008, 23:46
I agree with scthakur, maybe somehow goes wrong with answer.

I think n represents total terms from x to y, not the value of y.

I'm not sure if it's correct.

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Re: Please help # 1/4   [#permalink] 29 Oct 2008, 23:46
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The sum of the even numbers between 1 and n is 79*80, where

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