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Question Number 18448 by ankit kumar sharma last updated on 21/Jul/17

If  a, b, c, d are in GP and  a^x = b^y = c^z = d^u ,  then x, y, z, u are in

$$\mathrm{If}\:\:{a},\:{b},\:{c},\:{d}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{and}\:\:{a}^{{x}} =\:{b}^{{y}} =\:{c}^{{z}} =\:{d}^{{u}} , \\ $$$$\mathrm{then}\:{x},\:{y},\:{z},\:{u}\:\mathrm{are}\:\mathrm{in} \\ $$

Answered by Tinkutara last updated on 21/Jul/17

b^2  = ac, c^2  = bd  a = k^(1/x) , b = k^(1/y) , c = k^(1/z) , d = k^(1/u)   k^(2/y)  = k^((1/x) + (1/z)) , k^(2/z)  = k^((1/y) + (1/u))   ∴ x, y, z and u are in H.P.

$${b}^{\mathrm{2}} \:=\:{ac},\:{c}^{\mathrm{2}} \:=\:{bd} \\ $$$${a}\:=\:{k}^{\frac{\mathrm{1}}{{x}}} ,\:{b}\:=\:{k}^{\frac{\mathrm{1}}{{y}}} ,\:{c}\:=\:{k}^{\frac{\mathrm{1}}{{z}}} ,\:{d}\:=\:{k}^{\frac{\mathrm{1}}{{u}}} \\ $$$${k}^{\frac{\mathrm{2}}{{y}}} \:=\:{k}^{\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{z}}} ,\:{k}^{\frac{\mathrm{2}}{{z}}} \:=\:{k}^{\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{u}}} \\ $$$$\therefore\:{x},\:{y},\:{z}\:\mathrm{and}\:{u}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

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