Question

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Answer 1

$\bigstar$ Correct Question :

$\star$ $\bold{a^x=b^y=c^z,b^2=ac}$ . Then show that  $\bf{\frac{1}{x}+\frac{1}{z}=\frac{2}{y}}$

$\bigstar$ Solution :

$\star$ Let ,

$\bf{Let,\;a^x=b^y=c^z=k}\\\\ \implies\bf{a=k^{1/x},b=k^{1/y},c=k^{1/z}}$

$\star$ Now sub. these in b²=ac , we get ,

$\implies \bf{(k^{1/y})^2=k^{1/x}.k^{1/z}}\\\\\implies \bf{k^{2/y}=k^{1/x + 1/y}}$

$\star$ Bases are equal.So exponents must be equal.

$\implies\bf{\frac{2}{y}=\frac{1}{x}+\frac{1}{z}}$

Hence Proved

Answer 2

$\huge\mathcal{\underline{\underline{\pink{ANSWER}}}}$

GIVEN:-

• $a^x=b^y=c^z$

• $b^2=ac$

TO PROVE:-

• $\huge{\frac{1}{x}+\frac{1}{z}=\frac{2}{y}}$

Let,

$a^x=b^y=c^z$ =k

So,

$a=k^\frac{1}{x}$

$b=k^\frac{1}{y}$

$c=k^\frac{1}{z}$

Atq

$\huge{{k^\frac{1}{y}^2}={k^\frac{1}{x}}×{k^\frac{1}{z}}}$

Now, Taking the exponent as bases are same.

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

Hence proved.