If a, b, c are in AP and x, y, z are in GP then prove that the value of

Solution:

To prove: $x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1 \ldots$ (i)

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

$\Rightarrow 2 b=a+c \ldots$ (ii)

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

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

$\Rightarrow x=y^{2} / z$

Substitute this value of x in equation (i),we get

L.H.S =

$\Rightarrow\left(\frac{\mathrm{y}^{2}}{\mathrm{z}}\right)^{\mathrm{b}-\mathrm{c}} \times \mathrm{y}^{\mathrm{c}-\mathrm{a}} \times \mathrm{z}^{\mathrm{a}-\mathrm{b}}$

$\Rightarrow \mathrm{y}^{2(\mathrm{~b}-\mathrm{c})+\mathrm{c}-\mathrm{a}} \cdot \mathrm{z}^{\mathrm{a}-\mathrm{b}-(\mathrm{b}-\mathrm{c})}$

$\Rightarrow \mathrm{y}^{2 \mathrm{~b}-2 \mathrm{c}+\mathrm{c}-\mathrm{a}} \cdot \mathrm{z}^{\mathrm{a}+\mathrm{c}-\mathrm{b}-\mathrm{b}}$

$\Rightarrow \mathrm{y}^{2 \mathrm{~b}-\mathrm{c}-\mathrm{a}} \cdot \mathrm{z}^{\mathrm{a}+\mathrm{c}-2 \mathrm{~b}}$

$\Rightarrow y^{0} \cdot z^{0} \ldots($ Using equation (i))

= 1 = R.H.S

Hence, proved that. If $a, b, c$ are in $A P$ and $x, y, z$ are in $G P$ then $x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1$