If the terms of a G.P. are a, b and c, respectively.


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 $a^{a-r} b^{t-p} c^{p-q}=1$


Let A be the first term and R be the common ratio of the G.P.

According to the given information,

$A R^{p-1}=a$

$A R^{q-1}=b$

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

$a^{q-r} b^{r-p} c^{p-q}$

$=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}$

$=A q^{-r+r-p+p-q} \times R^{(p r-p r-q+r)+(r q-r+p-p q)+(p r-p-q r+q)}$

$=A^{0} \times R^{0}$


Thus, the given result is proved.

Leave a comment