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Math Help - Difficult arithmatic progression question

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    Difficult arithmatic progression question

    If m times the mth term of an AP is equal to n times its nth term, find the value of the
    (m + n)th term.

    well, after expanding and factorizing I stuck with dividing the whole thing by m-n/...
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by ice_syncer View Post
    If m times the mth term of an AP is equal to n times its nth term, find the value of the
    (m + n)th term.

    well, after expanding and factorizing I stuck with dividing the whole thing by m-n/...
    Well, I think you're on the right track ^^

    a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr
    a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr

    where a_0 is the first term of the sequence and r the constant progression.


    The text says that m \cdot a_m=n \cdot a_n

    m(a_0+mr)=n(a_0+nr)

    m \cdot a_0+m^2r-n \cdot a_0-n^2r=0

    r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0

    Factoring by m-n :

    (m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0

    Conclude ^^
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,


    Well, I think you're on the right track ^^

    a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr
    a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr

    where a_0 is the first term of the sequence and r the constant progression.


    The text says that m \cdot a_m=n \cdot a_n

    m(a_0+mr)=n(a_0+nr)

    m \cdot a_0+m^2r-n \cdot a_0-n^2r=0

    r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0

    Factoring by m-n :
    (m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0

    Conclude ^^
    Hey uh.... I was taught that the nth term is
    a1 + (n-1)d = nth term
    where a1 is the first term of AP
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  4. #4
    Moo
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    Quote Originally Posted by ice_syncer View Post
    Hey uh.... I was taught that the nth term is
    a1 + (n-1)d = nth term
    where a1 is the first term of AP
    It's the same. See, if you start at 0 instead of 1, you will have n instead of n-1.

    Ok, if you want to do this way...

    m[a_1+(m-1)r]=n[a_1+(n-1)r]

    ma_1+m^2r-mr-na_1-n^2r+nr

    a_1(m-n)+r[m^2-m-n^2+n]=0

    a_1(m-n)+r[m^2-n^2-(m-n)]=0

    a_1{\color{red}(m-n)}+r[{\color{red}(m-n)}(m+n)-{\color{red}(m-n)}]=0

    Factor by m-n :

    {\color{red}(m-n)}[a_1+(m+n-1)r]=0

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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,


    Well, I think you're on the right track ^^

    a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr
    a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr

    where a_0 is the first term of the sequence and r the constant progression.


    The text says that m \cdot a_m=n \cdot a_n

    m(a_0+mr)=n(a_0+nr)

    m \cdot a_0+m^2r-n \cdot a_0-n^2r=0

    r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0

    Factoring by m-n :

    (m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0

    Conclude ^^
    ok I got it the last factorization step was what helped me, thanks alot!
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