If thew first and sixth terms of geometric progessions are $$\frac{2}{3}$$ and 162 respectively then the $$8^{th}$$ term of the progression is

let the first term is a and common ratio is r then

$$T_{1}$$= {a}$$\times{r}^{1-1}$$ = a =  $$\frac{2}{3}$$                equation 1

$$T_{6}$$= {a}$$\times{r}^{6-1}$$ = {a}$$\times{r}^{5}$$ = 162      equation 2

equation 1 \equation 2

$$\frac{2}{3}$$\162 = 1\r^5

243 = r^5

3 = r

r = 3 

put the value of a and  r in equation following equation 

$$T_{8}$$= {a}$$\times{r}^{8-1}$$ =  $$\frac{2}{3}$$ $$\times{3}^{7}$$ 
                             $$T_{8}$$         =   $$\frac{2}{3}$$ $$\times243$$ $$\times3$$ $$\times3$$                                 

                              $$T_{8}$$        = 1458     Answer