Question

Find the 4th term from the end of the G. P. $\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{18}$, $\frac{1}{54}$ ,.....$\frac{1}{4374}$

Answer 1

$GP=\frac{1}{2},\frac{1}{6},\frac{1}{18},\frac{1}{54}.........\frac{1}{4374}$

$a=\frac{1}{2},$ $r=\frac{1/6}{1/2}=\frac{1}{3}$

$T_n=a{r}^{n-1}=\frac{1}{4374}\Rightarrow \frac{1}{2}(\frac{1}{3}{)}^{n-1}=\frac{1}{4374}$

$(\frac{1}{3}{)}^{n-1}=\frac{1}{2187}\Rightarrow (\frac{1}{3}{)}^{n-1}=(\frac{1}{3}{)}^7$

On compairig the powers $\Rightarrow n-1=7$

$\Rightarrow n=8$

$4^{th}$ term from end = $8-4+1=5^{th}$ term from beginning

$T_5=a{r}^4=\frac{1}{2}(\frac{1}{3}{)}^4=\frac{1}{162}$