Question

If $x,y,z$ be respectively the $p^{th}$, $q^{th}$, and $r^{th}$ terms of a $G.P$, then prove that
$(q-r)\log x+(r-p)\log y+(p-q)\log z=0$

Answer 1

$T_p=A{R}^{p-1}=x$, $T_q=A{R}^{q-1}=y$, $T_r=A{R}^{r-1}=z$
$\therefore \log x=\log A+(p-1)\log R$
$\log y=\log A+(q-1)\log R$
$\log z=\log A+(r-1)\log R$
Multiply $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ by $q-r$, $r-q$ and $p-q$ respectively and add.
$\therefore (q-r)\log x+(r-p)\log y+(p-q)\log z$
$=0\log A+0\log R=0$
$\because \Sigma (q-r)=0$, $\Sigma (p-1)(q-r)=0$