The sum of 100 terms of the series 0.9 + 0.09 | vedclass

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(A) The given series is $0.9 + 0.09 + 0.009 + \dots$ up to $100$ terms.This is a Geometric Progression $(G.P.)$ where the first term $a = 0.9 = \frac{

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$1 - \left( \frac{1}{10} \right)^{100}$

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(A) The given series is $0.9 + 0.09 + 0.009 + \dots$ up to $100$ terms.This is a Geometric Progression $(G.P.)$ where the first term $a = 0.9 = \frac{9}{10}$ and the common ratio $r = \frac{0.09}{0.9} = 0.1 = \frac{1}{10}$.The sum of the first $n$ terms of a $G.P.$ is given by the formula $S_n = a \frac{1 - r^n}{1 - r}$.Substituting the values $a = \frac{9}{10}$,$r = \frac{1}{10}$,and $n = 100$:$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{1 - \frac{1}{10}} \right)$$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{\frac{9}{10}} \right)$$S_{100} = 1 - \left( \frac{1}{10} \right)^{100}$.

The sum of $100$ terms of the series $0.9 + 0.09 + 0.009 + \dots$ will be

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