Question

If $x,y,z$ are in $G.P.$ and $x\ne y\ne z,$ when $x,y,z$ satisfy the condition $x^n+y^n=z^n,$ then

• A. ${\left(\frac{y}{z}\right)}^n={\left(\frac{x}{z}\right)}^{n/2}$
• B. ${\left(\frac{y}{z}\right)}^n=\frac{\sqrt{5}-1}{2}$
• C. ${\left(\sqrt{\frac{x}{z}}\right)}^n=\frac{\sqrt{5}-1}{2}$
• D. All of these

Answer 1

The correct option is D All of these

Given $x,y,z$ are in $G.P.$

$\therefore \frac{y}{x}=\frac{z}{y}\Rightarrow \frac{y^2}{z}=x$

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

$\Rightarrow \left(\frac{y}{z}\right)={\left(\frac{x}{z}\right)}^{1/2}\Rightarrow {\left(\frac{y}{z}\right)}^n={\left(\frac{x}{z}\right)}^{n/2}$

$[$ Hence (a) is coorrect.$]$

Again $x^n+y^n=z^n$

$\Rightarrow {\left(\frac{x}{z}\right)}^n+{\left(\frac{y}{z}\right)}^n=1\Rightarrow {\left(\frac{y}{z}\right)}^{2n}+{\left(\frac{y}{z}\right)}^n=1$

$\Rightarrow R^2+R-1=0$

Let $R={\left(\frac{y}{z}\right)}^n\Rightarrow R=\frac{-1\pm \sqrt{5}}{2}$

Now let $R=\frac{\sqrt{5}-1}{2}$

${\left(\frac{y}{z}\right)}^n=\frac{\sqrt{5}-1}{2}$ [(b) is correct.]

Again ${\left(\frac{y}{z}\right)}^n={\left(\frac{x}{z}\right)}^{n/2}=\frac{\sqrt{5}-1}{2}$ $[\therefore$ (c) is also correct$]$