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Math Help - G.P

  1. #1
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    Unhappy G.P

    If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0
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  2. #2
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    Quote Originally Posted by matsci0000 View Post
    If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0
    Hi

    x,y,z being respectively pth, qth and rth terms of a G.P (ratio k) you can write

    y = x \: k^{q-p} and z = x \: k^{r-p}
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  3. #3
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    Verify

    Can anyone check whether the question is correct or not ?

    I am getting a different solution.
    My working:

    " _ " sign has been used to denote subscript.

    Given:
    a_p=x a_q=y a_r=z

    Let 'a' be the first term and R be the common ratio of the G.P.
    aR^(p-1)=x aR^(q-1)=y aR^(r-1)=z
    logx=log a+(p-1)log R ------------- (1)
    logy=log a+(q-1)log R -------------(2)
    logz=log a+(r-1)log R ---------------(3)


    To prove that :
    (q-r)logx+(r-p)log y+(p-q)log z=0
    on solving LHS
    I obtained 2 log 0
    but log 0 is undefined.
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  4. #4
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    You should obtain
    (q-r)log a+(r-p)log a+(p-q)log a + [(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]log R=0
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  5. #5
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    Post solution to the question on G.P.

    Quote Originally Posted by matsci0000 View Post
    If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0
    nth term of G.P = aR ^ (n-1)
    a is the first term
    R is the common difference

    x = pth term y=qth term z = rth term
    x = aR^(p-1) y = aR^(q-1) z = aR^(r-1)
    log x = (p-1) log aR
    log y = (q-1) log aR
    log z = (r-1) log aR

    (q-r) log x + (r-p) log y + (p-q) log z

    = (q-r) (p-1) log aR + (r-p) (q-1) log aR + (p-q) (r-1) log aR

    = log aR [ (q-r)(p-1) + (r-p)(q-1) + (p-q)(r-1) ]
    =log aR [ qp - q -rp + r + rq - r -pq + p + pr -p -qr +q ]
    = log aR [0]
    = 0
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