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## Difficult math Q#1

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abby_g Junior | Next Rank: 30 Posts
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#### Difficult math Q#1

Sun Oct 01, 2006 8:35 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...

I think the answer is 159.

we have the series between 1 to n as 2,4,6,8,...,n-1

so take 2 common and we have
2[1,2,3,...,(n-1)/2]
the sum = 2[ (n-1)/2 * {(n-1)/2 + 1)]/2
= (n-1)(n+1)/4
equating on bothsides (n-1)(n+1)/4 = 79 *80
(n-1)(n+1) = 158 * 160
Hence n=159

The original method mistooks n(number of terms) for n (of 1 to n ) where n is number and not number of terms. Can anybody correct me please.

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gamemaster Senior | Next Rank: 100 Posts
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03 Sep 2006
Posted:
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Sun Oct 01, 2006 11:21 am
a quick way to solve this, is like this:

Lets say you have series of consequtive numbers:

A1 A2 A3 A4 ..... An

just to make things simple, lets say n is even (although it doesnt really matter)

its obvious that A1 + An = A2 + An-1 = A3 + An-2 and so on ... this is the definition of arithmetic progression

so you have n/2 couples with the same sum which is A1+An, the sum of the series will be n/2(An+A1) - Disclaimer: just to be accurate, im referring to n as the number of elements in the series, A1 doesnt necessarily has to be 1, and in this context 'n' is not necessarily the 'n' in An - i can clarify it if its not clear
i assume you'd understand what i mean ...

now for our private case:

our series is :

2 4 6 ..... n

so we actually have n/4 couples with a sum n+2 ! (n/4 because we first divide by 2 by creating a couple, but we have n/2 elements in the first place)

so: n/4 (n+2) = 79*80 => n = 158

This explenation was long, but if you remember it, you can solve these kind of questions in a flink of an eye

abby_g Junior | Next Rank: 30 Posts
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Sun Oct 01, 2006 6:06 pm
so we actually have n/4 couples with a sum n+2 !

I did not get why sum is n+2!. I'm convinced with n/4 couples.

By the way cool approach. thanks.

abby_g Junior | Next Rank: 30 Posts
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Sun Oct 01, 2006 6:10 pm
Hey got it 'n' is the number of terms, then n/2 couples and the sum will be n/2 (A1+ An) which in this case n/4 * (n+2). Great.

thanks a lot. THat was real quickie.

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