Question

Find the sum of the number of terms in the geometric progressions in

$0.15,0.015,0.0015,....20$

Answer 1

Here $a=0.15,r=\frac{0.015}{0.15}=0.1$ and $n=20$

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

$S_{20}=\frac{0.15(1-{0.1}^{20})}{1-0.1}$

$=\frac{0.15(1-{0.1}^{20})}{0.9}$

$=\frac{15(1-{0.1}^{20})}{90}$

$=\frac{11(1-{0.1}^{20})}{6}$