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Difficult math Q#1

This topic has 3 member replies
abby_g Junior | Next Rank: 30 Posts Default Avatar
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Difficult math Q#1

Post 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]
    Arrow the sum = 2[ (n-1)/2 * {(n-1)/2 + 1)]/2
    Arrow = (n-1)(n+1)/4
    Arrow equating on bothsides (n-1)(n+1)/4 = 79 *80
    Arrow (n-1)(n+1) = 158 * 160
    Arrow 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 Default Avatar
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    Post 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 Default Avatar
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    Post 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!. Question I'm convinced with n/4 couples.

    By the way cool approach. thanks.

    abby_g Junior | Next Rank: 30 Posts Default Avatar
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    Post Sun Oct 01, 2006 6:10 pm
    Idea 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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