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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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