Question

If a, b, c are in AP and x, y, z are in GP then prove that the value of $x^{b-c}\cdot y^{c-a}\cdot z^{a-b}$ is $1$.

Answer 1

Given,

$a,b,c$ are in AP$\Rightarrow 2b=a+c$

$x,y,z$ are in GP$\Rightarrow y^2=xz$

$\therefore x^{(b-c)}\cdot y^{(c-a)}\cdot z^{(a-b)}$

$=x^{(b-c)}\cdot (\sqrt{xz}{)}^{(c-a)}\cdot z^{(a-b)}$ $[\because y=\sqrt{xz}]$

$=x^{(b-c)}\cdot x^{\frac{1}{2}(c-a)}\cdot z^{\frac{1}{2}(c-a)}\cdot z^{(a-b)}$

$=x^{(b-c)+\frac{1}{2}(c-a)}\cdot z^{\frac{1}{2}(c-a)+(a-b)}$

$=x^{\frac{1}{2}\{2b-(a+c\left)\right\}}\cdot z^{\frac{1}{2}\{c+a-2b\}}$
$=x^0\times z^0=1$.