Question

The (m + n)th and the (m - n)th terms of a GP are p and q respectively. Show that the mth and the nth terms of the GP are $\sqrt{pq}$ and ${\left(\frac{q}{p}\right)}^{\left(\frac{m}{2n}\right)}$

Answer 1

Let a be the first term and r be the common ratio of the given GP.

Then,

$T_{m+n}=p$ and $T_{m-n}=q$

$\Rightarrow a{r}^{(m+n-1)}=p$ . . .(i) and $\Rightarrow a{r}^{(m-n-1)}=q$ . . . (ii)

$\Rightarrow \frac{a{r}^{(m+n-1)}}{a{r}^{m-n-1}}=\frac{p}{q}$

$\Rightarrow r^{\left[\right(m+n-1)-(m-n-1\left)\right]}=\frac{p}{q}$

$\Rightarrow r^{2n}=\frac{p}{q}\Rightarrow r={\left(\frac{p}{q}\right)}^{\frac{1}{2n}}$

$\Rightarrow \frac{1}{r}={\left(\frac{q}{p}\right)}^{\frac{1}{2n}}$ . . . (iii)

$\therefore T_m=a{r}^{(m-1)}$

$=a{r}^{(m+n-1)}\times {\left(\frac{1}{r}\right)}^n$

$=p\times {\left\{{\left(\frac{q}{p}\right)}^{\frac{1}{2n}}\right\}}^n$ [using (i) and (iii)]

$=p\times {\left(\frac{q}{p}\right)}^{(\frac{1}{2n}\times n)}=p\times {\left(\frac{q}{p}\right)}^{\frac{1}{2}}=\sqrt{pq}$

And, $T_n=a{r}^{(n-1)}$

$=a{r}^{(m+n-1)}\times {\left(\frac{1}{r}\right)}^m$

$=p\times {\left\{{\left(\frac{q}{p}\right)}^{\frac{1}{2n}}\right\}}^m=p\times {\left(\frac{q}{p}\right)}^{\frac{m}{2n}}$ [using (i) and (iii)]

Hence, $T_m=\sqrt{pq}$ and $T_n=p\times {\left(\frac{q}{p}\right)}^{\frac{m}{2n}}$