Question:
If the $p^{b \prime}, q^{t h}$ and $r^{\prime \prime}$ terms of a G.P. are $a, b$ and $c$, respectively. Prove that $a^{a-r} b^{t-p} c^{p-q}=1$
Solution:
Let A be the first term and R be the common ratio of the G.P.
According to the given information,
$A R^{p-1}=a$
$A R^{q-1}=b$
$A R^{r-1}=c$
$a^{q-r} b^{r-p} c^{p-q}$
$=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}$
$=A q^{-r+r-p+p-q} \times R^{(p r-p r-q+r)+(r q-r+p-p q)+(p r-p-q r+q)}$
$=A^{0} \times R^{0}$
$=1$
Thus, the given result is proved.