A cool geometrical progression!!

The series converges by the geometric series test. Therefore we can let $S = 0.9 + 0.03 + 0.001 + ...$ $S = 0.9 + \frac{1}{30}(0.9 + 0.03 + 0.001 + ...)$ $S = \frac{9}{10} + \frac{S}{30}$ $30\times S = 27 + S$ $29\times S = 27$ $S = \boxed{\frac{27}{29}}$

In this question,

$a = 0.9 = \frac{9}{10}$ ,

$r = \frac{.03}{0.9} = \frac{3}{90} = \frac{1}{30}$ (r<1)

sum of infinite terms of G.P. (S) = $\frac{a}{1-r}$

$S= \frac{\frac{9}{10}}{1 - \frac{1}{30}}$

$S = \frac{\frac{9}{10}}{\frac{29}{30}}$

$S = \frac{9}{10} × \frac{30}{29}$

$\boxed{ S = \frac{27}{29}}$

The series would go something like :

0.9+0.03+0.001+0.0001+0.00001+0.000001..........

so the sum would be :

0.9311111111111111111111.................

or 0.931 (approximately)

Only the option 27/29 gives this value.