Find the sum to 20 terms in the geometric progression
0.15, 0.015, 0.0015 ...

Here a = 0.15 and r = $\frac{0.015}{0.15}=\frac{1}{10}$

We know that

$S_n=\frac{a(1-r^n)}{1-r}whenr<1$

$S_{20}=\frac{0.15[1-{\left(\frac{1}{10}\right)}^{20}]}{1-\frac{1}{10}}$

= $\frac{15}{100}\times \frac{10}{9}[1-(0.1{)}^{20}]$

= $\frac{1}{6}[1-(0.1{)}^{20}]$