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Question Number 3617 by Rasheed Soomro last updated on 16/Dec/15

If  the G.M. between two numbers   a , b   is  G and the two A.M.′s between  them are p  and  q , then prove that  G^2 =(2p−q)(2q−p).

$${If}\:\:{the}\:{G}.{M}.\:{between}\:{two}\:{numbers}\: \\ $$$${a}\:,\:{b}\:\:\:{is}\:\:{G}\:{and}\:{the}\:{two}\:{A}.{M}.'{s}\:{between} \\ $$$${them}\:{are}\:{p}\:\:{and}\:\:{q}\:,\:{then}\:{prove}\:{that} \\ $$$${G}^{\mathrm{2}} =\left(\mathrm{2}{p}−{q}\right)\left(\mathrm{2}{q}−{p}\right). \\ $$

Commented by Yozzii last updated on 16/Dec/15

There are multiple definitions of the  A.M between a and b?

$${There}\:{are}\:{multiple}\:{definitions}\:{of}\:{the} \\ $$$${A}.{M}\:{between}\:{a}\:{and}\:{b}? \\ $$

Commented by prakash jain last updated on 16/Dec/15

AP: a, a+d, a+2d, b=a+3d  p=a+d,q=a+2d  2p−q=2a+2d−a+2d=a  2q−p=2(a+2d)−(a+d)=a+3d=b  (2p−q)(2q−p)=ab=G^2

$$\mathrm{AP}:\:{a},\:{a}+{d},\:{a}+\mathrm{2}{d},\:{b}={a}+\mathrm{3}{d} \\ $$$${p}={a}+{d},{q}={a}+\mathrm{2}{d} \\ $$$$\mathrm{2}{p}−{q}=\mathrm{2}{a}+\mathrm{2}{d}−{a}+\mathrm{2}{d}={a} \\ $$$$\mathrm{2}{q}−{p}=\mathrm{2}\left({a}+\mathrm{2}{d}\right)−\left({a}+{d}\right)={a}+\mathrm{3}{d}={b} \\ $$$$\left(\mathrm{2}{p}−{q}\right)\left(\mathrm{2}{q}−{p}\right)={ab}={G}^{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 16/Dec/15

n AMs x_i  between a and b means  a,x_1 ,x_2 ,...,x_n ,b is an AP.

$${n}\:\mathrm{AM}{s}\:{x}_{{i}} \:\mathrm{between}\:{a}\:\mathrm{and}\:{b}\:\mathrm{means} \\ $$$${a},{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,...,{x}_{{n}} ,{b}\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}. \\ $$

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