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

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Question

(A) The given series 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.

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given series 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 $S_n = a \frac{1 - r^n}{1 - r}$.For $n = 100$,we have:$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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