Question:
Which term of the progression 0.004, 0.02, 0.1, ... is 12.5?
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 $12.5$