# 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 $\displaystyle 12\ 288$.
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