Question

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

Answer 1

Given,

$a=a{r}^{p-1}$

$b=b{r}^{q-1}$

$c=c{r}^{r-1}$

$\Rightarrow a^{q-r}=[a{r}^{p-1}{]}^{q-r}$

$\Rightarrow b^{r-p}=[a{r}^{q-1}{]}^{r-p}$

$\Rightarrow c^{p-q}=[a{r}^{r-1}{]}^{p-q}$

$\therefore a^{q-r}\times b^{r-p}\times c^{p-q}=[a{r}^{p-1}{]}^{q-r}\times [a{r}^{q-1}{]}^{r-p}\times [a{r}^{r-1}{]}^{p-q}$

$=a^{(p+q+r-p-q-r)}{r}^{(pq+qr+rp+p+r+q-qp-rp-rq-p-q-r)}$

$=a^0{r}^0$

$\therefore a^{q-r}\times b^{r-p}\times c^{p-q}=1$

Hence proved