Question

If the mth,nth and pth terms of an A.P. and G.P. be equal and be respectively $x,y$ and $z,$ then prove that $x^{y-z}\cdot y^{z-x}\cdot z^{x-y}=1$

Answer 1

By given conditions,
$x=a+(m-1)d=A{R}^{m-1}for{T}_m$
$y=a+(n-1)d=A{R}^{n-1}for{T}_n$
$z=a+(p-1)d=A{R}^{p-1}for{T}_p$
$\therefore y-z=(n-p)d,z-x=(p-m)d,$
$x-y=(m-n)d$ ....(1)
$\therefore x^{y-z}.y^{z-x}.z^{x-y}$
$=\left(A{R}^{m-1}{)}^{(n-p)d}\right(A{R}^{n-1}{)}^{(p-m)d}(A{R}^{p-1}{)}^{(m-n)d}$
$=A^{\circ }.R^{\circ }=1$
$\because$$(n-p+p-m+m-n)d=0$ and $d\left[\right(m-1\left)\right(n-p)+(n-1\left)\right(p-m)+(p-1\left)\right(m-n\left)\right]=0$