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Thread: Question on GP, AP and HP

  1. #1
    Junior Member
    Jul 2009

    Unhappy Question on GP, AP and HP

    If a,b,c are in AP ,b,c,d are in GP and c,d,e are in HP ; the prove that a,c,e are in GP.
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  2. #2
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK
    My approach would be to start with the given relationships:

    $\displaystyle b-a = c-b$ for the A.P. property

    $\displaystyle \frac c b = \frac d c$ for the G.P. property

    $\displaystyle \frac 1 d - \frac 1 c = \frac 1 e - \frac 1 d$ for the H.P property.

    Tidy up as far as possible, then substitute for $\displaystyle b$ and $\displaystyle d$ to get an equation demonstrating the result, which should be reducible to:

    $\displaystyle \frac c a = \frac e c$

    I've just tried it on paper but haven't got the result I was expecting - I probably made a mistake in my algebra (easy to do). Try it out, see what you get, post your working on here and see what you get. (I won't be on line again today as from 20 minutes from now till about 12 hours time, unfortunately.)

    Edit: Yes, I've tried it again, and the above approach does work.
    Last edited by Matt Westwood; Aug 31st 2009 at 10:48 PM. Reason: added a few bits
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
    Dec 2007
    IISc, Bangalore
    From the given data, $\displaystyle 2b = a+c, c^2 = bd, \frac{2}{d} = \frac1{c} + \frac1{e}$. Also observe that the last equation ensures that d,c or e cant be 0.

    Now $\displaystyle c^2 = bd = \frac{a+c}2 \cdot \dfrac2{\frac1{c} + \frac1{e}}$. Simplifying the right hand side of the equation,

    $\displaystyle c^2 = \frac{(a+c)ce}{c+e}$.

    Using c is not 0 and cross multiplying the above equation,

    $\displaystyle c(c+e) = (a+c)e \implies c^2 + ec = ae + ce \implies \boxed{c^2 = ae}$
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