Solve this


If $a, b, c$ are in GP and $a^{1 / x}=b^{1 / y}=c^{1 / z}$ then prove that $x, y, z$ are in AP.



It is given that

$a^{1 / x}=b^{1 / y}=c^{1 / z}$

Let $a^{1 / x}=b^{1 / y}=c^{1 / z}=k$

$\Rightarrow a^{1 / x}=k$

$\Rightarrow\left(a^{1 / x}\right)^{x}=k^{x} \ldots($ Taking power of $x$ on both sides. $)$

$\Rightarrow a^{1 / x \times x}=k^{x}$

$\Rightarrow a=k^{x}$

Similarly $b=k^{y}$

And $c=k^{z}$

It is given that a,b,c are in G.P.

$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$

Substituting values of a,b,c calculated above, we get:

$\Rightarrow\left(\mathrm{k}^{\mathrm{y}}\right)^{2}=\mathrm{k}^{\mathrm{x}} \mathrm{k}^{\mathrm{z}}$

$\Rightarrow \mathrm{k}^{2 \mathrm{y}}=\mathrm{k}^{\mathrm{x}+\mathrm{z}}$

Comparing the powers we get,

$2 y=x+z$

Which is the required condition for x,y,z to be in A.P.

Hence, proved that x,y,z, are in A.P.



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