Which term of the progression 0.004, 0.02, 0.1,

Solution:

We have;

$\frac{a_{2}}{a_{1}}=\frac{0.02}{0.004}=5, \frac{a_{3}}{a_{2}}=\frac{0.1}{0.02}=5$

$\Rightarrow \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=5$

The given progression is a G.P. whose first term, $a$ is $0.004$ and common ratio, $r$ is 5 .

Let the $n$th term be $12.5$.

$\therefore a_{n}=12.5$

$\Rightarrow a r^{n-1}=12.5$

$\Rightarrow(0.004)(5)^{n-1}=12.5$

$\Rightarrow(5)^{n-1}=\frac{12.5}{0.004}$

$\Rightarrow(5)^{n-1}=3125$

$\Rightarrow(5)^{n-1}=(5)^{5}$

Comparing the power of both the sides

$\Rightarrow n-1=5$

$\Rightarrow n=6$

Thus, 6 th term of the given G.P. is $12.5$