Q. The sum of 100 terms of the series $0.9 + 0.09 + 0.009\ldots$ will be
A) $1 - \left( \frac{1}{10} \right)^{100}$
B) $1 + \left( \frac{1}{10} \right)^{100}$
C) $1 - \left( \frac{1}{10} \right)^{106}$
D) $1 + \left( \frac{1}{10} \right)^{10}$
Explanation
(a)
Here, $a = 0.9 = \frac{9}{10}$ and $r = \frac{1}{10} = 0.1$
$\therefore\ S_{100} = a\left( \frac{1 - r^{100}}{1 - r} \right)$
$= \frac{9}{10}\left( \frac{1 - \frac{1}{10^{100}}}{1 - \frac{1}{10}} \right) = 1 - \left( \frac{1}{10^{100}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.