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if  a,b, c  are in G.P and a^{x}=b^{y}=c^{z} then

 

  • Option 1)

    \frac{1}{x}+\frac{1}{z}=\frac{2}{y}

  • Option 2)

    \frac{1}{x}+\frac{1}{z}=-\frac{2}{y}

  • Option 3)

    \frac{1}{x}+\frac{1}{y}=\frac{2}{z}

  • Option 4)

    \frac{1}{x}+\frac{1}{y}=-\frac{2}{z}

 

Answers (1)

best_answer

As we learnt in 

Geometric mean of two numbers (GM) -

GM= \sqrt{ab}

- wherein

 

 a,b,c are in GP

So b^{2}=ac\\*\\*a^{x}=b^{y}=c^{z}=K(Let's\ say)\\*\\* a=K^{\frac{1}{x}};b=K^{\frac{1}{y}};c=K^{\frac{1}{z}}\\*\\*b^{2}=ac\\*\\*= > K^{\frac{2}{y}}=K^{\frac{1}{x}+\frac{1}{z}}


Option 1)

\frac{1}{x}+\frac{1}{z}=\frac{2}{y}

Correct

Option 2)

\frac{1}{x}+\frac{1}{z}=-\frac{2}{y}

Incorrect

Option 3)

\frac{1}{x}+\frac{1}{y}=\frac{2}{z}

Incorrect

Option 4)

\frac{1}{x}+\frac{1}{y}=-\frac{2}{z}

Incorrect

Posted by

Aadil

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