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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- Aug 31st 2009, 08:16 PM #1

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- Aug 31st 2009, 10:34 PM #2
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.

- Sep 1st 2009, 12:54 AM #3
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,

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