Question

If the $p^{th},q^{th}$ and $r^{th}$ terms of a G.P. are a, b and c, respectively. Prove that $a^{q-r}{b}^{r-p}{c}^{p-q}=1$.

Answer 1

$LetfirsttermAandcommondifference\left(R\right)\begin{array}{c}{t}_p=a\\ t_q=b\\ t_r=c\end{array}|G.P.t_p=A{R}^{p-1}=a{t}_q=A{R}^{q-1}=b{t}_r=A{R}^{r-1}=cthen{a}^{q-r}.b^{r-p}.c^{p-q}{\left\{A{R}^{p-1}\right\}}^{q-r}{\left\{A{R}^{q-1}\right\}}^{r-p}{\left\{A{R}^{r-1}\right\}}^{p-q}{A}^{\left(q-r+r-p+p-a\right)}.R^{\left\{\left(p-1\right)\left(q-1\right)+\left(q-1\right)\left(r-p\right)+\left(r-1\right)\left(p-q\right)\right\}}{A}^{\circ }.R^{\circ }=1\therefore a^{q-r}.b^{r-p}.c^{p-q}=1Proved.$