Arithmetic progression & Geometric progression

Question

If the $m$-th, $n$-th, and $p$-th terms of an A.p. and G.p. are equal and are $x,y,z$ respectively, prove that $x^{y-z}$. $y^{z-x}$. $z^{x-y}= 1$. To solve this question what I did is simply kept values of $x,y,z$ from G.P. (i.e. for $x = ar^{m-1}$ so on). Can you help me to solve this in an more interesting way.

Answer 1

Let $x=a+(m-1)d=br^{m-1}$ etc. where $a,d$ are the first term & the common difference respectively of the AP and $b,r$ are the first term & the common ratio respectively of the GP

$y-z=(n-p)d$

$$\prod_{\text{cyc}}x^{y-z}=\prod_{\text{cyc}}(br^{m-1})^{d(n-p)}=\left(\dfrac br\right)^{d\sum_{\text{cyc}}(n-p)}r^{d\sum_{\text{cyc}}m(n-p)}$$

Now $\sum_{\text{cyc}}(n-p)=0=\sum_{\text{cyc}}m(n-p)$

Answer 2

Let $d$ be the common difference of the A.P. and $r$ be the common ratio of the G.P. Then

$$\begin{cases} \displaystyle \frac{y}{x}=r^{n-m}=r^{(y-x)/d}\\ \displaystyle\frac{x}{z}=r^{m-p}=r^{(x-z)/d}\\ \displaystyle\frac{z}{y}=r^{p-n}=r^{(z-y)/d}\end{cases}$$

Therefore,

\begin{align} \left(\frac{y}{x}\right)^z\left(\frac{x}{z}\right)^y\left(\frac{z}{y}\right)^x&=r^{[z(y-x)+y(x-z)+x(z-y)]/d}\\ x^{y-z}y^{z-x}z^{x-y}&=1 \end{align}

Answer 3

Using the formulas: $a_n-a_m=d(n-m)$ (A.P.) and $b_n=b_m\cdot r^{n-m}$ (G.P.),

$$x^{y-z}y^{z-x}z^{x-y}=x^{d(n-p)}\cdot (xr^{n-m})^{d(p-m)}\cdot (xr^{p-m})^{d(m-n)}=x^0\cdot r^0=1.$$