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Question

If $$x_{1}, x_{2}, x_{3}$$.......... are ion $$G.P$$. of natural numbers such that $$x_{1}.x_{2}.x_{3}x_{4}=64$$ then $$x_{5}=4^{k}$$ find $$k$$


Solution

$$\begin{array}{l} { x_{ 1 } },{ x_{ 2 } },{ x_{ 3 } }\, \, are\, \, in\, \, G.P &  \\ \Rightarrow { x_{ 1 } }.{ x_{ 2 } }.{ x_{ 3 } }.{ x_{ 4 } }=64 &  \\ \Rightarrow { x_{ 1 } }.{ x_{ 2 } }.{ x_{ 3 } }.{ x_{ 4 } }.{ x_{ 5 } }={ 4^{ 3 } }{ .4^{ k } } &  \\ let\, \, \frac { 1 }{ { { r^{ 2 } } } } .\frac { 1 }{ r } .1,r,{ r^{ 2 } }\, are\, \, in\, \, G.P &  \\ Hence,\, \, \frac { 1 }{ { { r^{ 2 } } } } .\frac { 1 }{ r } .1.r.{ r^{ 2 } }={ 4^{ 3+k } } &  \\ \Rightarrow 1={ 4^{ 3+k } } & \left[ \begin{array}{l} { a^{ x } }=y \\ x=y \end{array} \right]  \\ \Rightarrow { 4^{ 0 } }={ 4^{ 3+k } } &  \\ \therefore 3+k=0 &  \\ k=-3 &  \end{array}$$

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