If $$6^{th}$$ term and $$13^{th}$$ term of a geometric progression are 24 and $$\frac{3}{16}$$ respectively, then the $$25^{th}$$ term is

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

$$T_{6}$$= {a}$$\times{r}^{5}$$ =24           equation 1                                                            

$$T_{13}$$= {a}$$\times{r}^{12}$$ =3\16     equation 2

equation 1 \equation 2

24\{3\16} = 1\r^7

128  = 1\r^7 

2 = 1\r 

r = 1\2 

put the value of r in equation 1 

a$$\times\frac{1}{2}^{5}$$ =24 

so we get a= 768  so the series is 

768, 384,192,96,48,24,12,6,3,3\2,3\2^2,.............. so the 25^{th} term is 

 
 $$\frac{3}{2^{16}}$$   answer