Question

In a $G.P$ if $T_1=l$ and $T_m=m$, then prove that its $n^{th}$ term is ${\left[\frac{l^{n-m}}{m^{n-1}}\right]}^{-1/(l-m)}$

Answer 1

$T_l=l$

$T_m=m$

$a{r}^{{}^{l-1}}=l$

$a{r}^{m-1}=m$ $\Rightarrow r^{(l-m)}=\frac{l}{m}$

$T_n=a{r}^{n-1}$

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

$={\left(\frac{l^{n-m}}{m^{n-1}}\right)}^{\frac{-1}{l-m}}$