Question

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

Answer 1

Formula,

$a_n=a{r}^{n-1}$

$a=a{r}^{p-1}\Rightarrow a^{q-r}=(a{r}^{p-1}{)}^{q-r}$........(1)

$b=a{r}^{q-1}\Rightarrow b^{r-p}=(b{r}^{q-1}{)}^{r-p}$........(2)

$c=a{r}^{r-1}\Rightarrow c^{p-q}=(c{r}^{r-1}{)}^{p-q}$........(3)

multiplying (1),(2) and (3), we get,

$a^{q-r}{b}^{r-p}{c}^{p-q}=\left(a{r}^{p-1}{)}^{q-r}\right(b{r}^{q-1}{)}^{r-p}(c{r}^{r-1}{)}^{p-q}$

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

$=a^0{r}^0$

$=1$

Hence proved.