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

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Manager
Joined: 16 Oct 2008
Posts: 54
The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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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=?
Manager
Joined: 16 Oct 2008
Posts: 54

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28 Oct 2008, 06:58
Director
Joined: 14 Aug 2007
Posts: 704

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28 Oct 2008, 07:09
Natasha0123 wrote:

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
SVP
Joined: 17 Jun 2008
Posts: 1504

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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.
Manager
Joined: 16 Oct 2008
Posts: 54

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

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