Question

The $r^{th}$, $s^{th}$ and $t^{th}$ terms of a certain $G.P$ are
$R,S$ and $T$ respectively. Prove that
$R^{s-t}{S}^{t-r}{T}^{r-s}=1$.

Answer 1

Let the common ratio be. taken as $k$ and $a$ be the first term.
$\therefore R=T_r=a{k}^{r-1}$
$\therefore R^{s-t}=a^{s-t}{k}^{(r-1)(s-t)}$ Similarly,
$S^{t-r}=a^{t-r}{k}^{(s-1)(t-r)}$
$T^{r-s}=a^{r-s}{k}^{(t-1)(r-s)}$
Multiplying the above three and knowing that
$A^m.A^n.A^p=A^{m+n+p}$
$\therefore R^{s-t}{S}^{t-r}{T}^{r-s}=a^0$. $k^0=1$.
$\because \Sigma (a-b)=0,\Sigma (a+\lambda )(b-c)=0$