Question

If the $p^{th},q^{th}$ and $r^{th}$ term of a GP are $a,b,c$ respectively, then $a^{q-r}.b^{r-p}.c^{p-q}$ is equal to

• A. $0$
• B. $1$
• C. $abc$
• D. $pqr$

Answer 1

The correct option is B

$1$

Explanation of the correct option.

Compute the required value.

Let $A$ be the first term of GP with common ration $R$.

The $n^{th}$ term $A_n=A{R}^{n-1}$

Given: The $p^{th},q^{th}$ and $r^{th}$ term of a GP are $a,b,c$.

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

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

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

So, $a^{q-r}·b^{r-p}·c^{p-q}={\left(A{R}^{p-1}\right)}^{q-r}{\left(A{R}^{q-1}\right)}^{r-p}{\left(A{R}^{r-1}\right)}^{p-q}$

$$\begin{gathered} =A^{\left(q-r+r-p+p-q\right)}{R}^{\left(p-1\right)\left(q-r\right)+\left(q-1\right)\left(r-p\right)+\left(r-1\right)\left(p-q\right)} \\ =A^0{R}^{pq-pr-q+r+qr-r-pq+p+pr-p-qr+q} \\ =A^0{R}^0 \\ =1 \end{gathered}$$

Therefore, $a^{q-r}.b^{r-p}.c^{p-q}=1$.

Hence option $\left(B\right)$ is the correct option.