Problem Study : Geometric Progression

Find the smallest  number in the progression 3, 12, 48, ... which is greater than 10 000.

Solution

          Given Sequence : $ \displaystyle 3, 12, 48, ...$

       $ \displaystyle \therefore \frac{{{{u}_{2}}}}{{{{u}_{1}}}}=\frac{{12}}{3}=4$

       $ \displaystyle \ \ \ \frac{{{{u}_{3}}}}{{{{u}_{2}}}}=\frac{{48}}{{12}}=4$

       $ \displaystyle \therefore \frac{{{{u}_{2}}}}{{{{u}_{1}}}}=\frac{{{{u}_{3}}}}{{{{u}_{2}}}}$

       Therefore the given sequence is a geometric progression with the first term $ \displaystyle 3$ and the commratio $ \displaystyle 4$.
     
       $ \displaystyle \therefore a=3$  and $ \displaystyle r=4$.

       Let the smallest number in the progression which is greater than $ \displaystyle 10\ 000$ be $ \displaystyle {{{u}_{n}}}$

      $ \displaystyle \therefore {{u}_{n}}>10\ 000$    

      $ \displaystyle \ \ a{{r}^{{n-1}}}>10\ 000$    

      $ \displaystyle \ 3({{4}^{{n-1}}})>10\ 000$    
 
      $ \displaystyle \ \ {{4}^{{n-1}}}>3333.33$

      But $ \displaystyle {{4}^{5}}=1024<3333 .33$ and $ \displaystyle {{4}^{6}}=4096>3333 .33$
 
     $ \displaystyle \therefore n-1=6$
 
     $ \displaystyle \therefore n=7$
 
     $ \displaystyle \therefore {{u}_{7}}=3({{4}^{6}})=12\ 288$
 
     Therefore the smallest number in the progression which is greater than $ \displaystyle 10\ 000$ is $ \displaystyle 12\ 288$.