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Math Help - Arithmetic Progression

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    Arithmetic Progression

    2 Questions:

    1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

    2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.

    Thank you for your help!
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  2. #2
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    Hi

    I don't know what the meaning of "special" in "special arithmetic progression is...
    Quote Originally Posted by Tangera View Post
    2 Questions:

    1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.
    Let r be the constant progression of this sequence.

    We have 731=663+2r because they're separated by 2 terms.

    So r=34

    Now, note that 731=34*21+17
    It all depends on where you want to start the sequence, but if it has to be the first positive term, it would be a_0=17 and M=21

    With these informations, I guess you can continue, can't you ?
    Last edited by Moo; June 12th 2008 at 12:56 AM. Reason: thanks down there ! :D
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    Quote Originally Posted by Moo View Post
    Hi

    I don't know what the meaning of "special" in "special arithmetic progression is...


    Let r be the constant progression of this sequence.

    We have 731=663+2r because they're separated by 2 terms.

    So r=34

    Now, note that 731=34*21+17
    It all depends on where you want to start the sequence, but if it has to be the first positive term, it would be a_0=17 and M=34 Shouldn't it be M = 21 ?

    With these informations, I guess you can continue, can't you ?
    ..
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    Thank you for your help! I would like to know if it is possible to work out that M = 22 without using the fact that 731 = 34 x 21 + 17...because that would involve a bit of trial and error right?

    Could you guide me to solve question 2 as well? Thank you!

    2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.
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    Hi

    Thank you for your help! I would like to know if it is possible to work out that M = 22 without using the fact that 731 = 34 x 21 + 17...because that would involve a bit of trial and error right?
    Not trial and error. It's just that you should know this :

    a_n=a_m+(n-m)r

    Here, n=M and m=0
    And n-m=21, because r=34.

    So M-0=21 --> M=21

    I'm sorry for the second problem, I can't really see it for the moment
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    Sorry! I still don't understand why n-m=21 because r=34...? Aren't there 2 variables in that same equation??

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    Quote Originally Posted by Tangera View Post
    Sorry! I still don't understand why n-m=21 because r=34...? Aren't there 2 variables in that same equation??

    Hmmm... "because" was written because I sort of identified the numbers. To resume :

    a_n=a_m+(n-m)r

    Here, we have : \underbrace{a_M}_{731}=\underbrace{a_0}_{17}+34 \times 21

    So it's clear that a_M=a_n, that is to say n=M
    and a_m=a_0, that is to say m=0

    There remain (n-m) and r to determine. r is the progression, so 34=r. Thus remains (n-m), which will be 21.
    we know from above that n-m=M-0.
    So M-0=21 ----> M=21


    I hope this clears out the problem...sorry for not being clear
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    Quote Originally Posted by Tangera View Post
    2 Questions:

    1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

    2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.

    Thank you for your help!
    We require additional information for problem 2. Or a least tell us what a "special" A.P is?

    To prove the data is insufficient for a fixed N, note that the A.P with a = 1 and d=2 gives N^2 as the sum of first N terms and thereby the sum of next N terms is 3N^2.

    However The A.P with a = N, d=0 gives the same sum of first N terms but the sum of next N terms is N^2.

    All I can prove is Sum of Next N terms = S + n^2d.
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