Question

If $(p+q)th$ term of a G.P be m and $(p-q)th$ term be $n$, then the $pth$ term will be

• A. $\frac{m}{n}$
• B. $mn$
• C. $\sqrt{\left(mn\right)}$
• D. $0$

Answer 1

The correct option is C

$\sqrt{\left(mn\right)}$

Explanation for the correct option

Step 1: Information required for the solution

Let $a$ be the first term of GP with a common ratio $r$

Then the last term of the GP will be $a{r}^{n-1}$

The ${\left(p+q\right)}^{\mathrm{th}}$ term will be,

$a{r}^{(p+q-1)}=m\dots \left(1\right)$

The ${\left(p-q\right)}^{\mathrm{th}}$ term will be,

$a{r}^{(p-q-1)}=n\dots \left(2\right)$
Step 2: Calculation for the correct option

From the multiplication of equation $\left(1\right)\text{and}\left(2\right)$, we have

$\begin{array}{rcl}nm& =& a{r}^{(p+q-1)}.\left(a{r}^{(p-q-1)}\right)\\ & =& a^2{r}^{(2p-2)}\\ & =& {\left(a{r}^{p-1}\right)}^2\end{array}$

From this we have

$\left(a{r}^{p-1}\right)=\sqrt{nm}$ $\left[\because a{r}^{p-1}=p^{th}term\right]$

Hence option (C) is the correct option.