Solution:
Using the formula,
$\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$
Given that,
$\begin{array}{l}
a=0.004 \\
r=t_{2} / t_{1}=(0.02 / 0.004) \\
=5 \\
T_{n}=12.5 \\
n=?
\end{array}$
$\mathrm{So}, \mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$
$\begin{array}{l}
12.5=(0.004)(5)^{n-1} \\
12.5 / 0.004=5^{n-1} \\
3000=5^{n-1} \\
5^{5}=5^{n-1} \\
5=n-1 \\
n=5+1 \\
=6
\end{array}$
Therefore, $6^{\text {th }}$ term of the progression $0.004,0.02,0.1, \ldots$ is $12.5$