The value of overline{0.037},where overline{0 | vedclass

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Question

(D) Let $x = 0.037037037...$ This is a repeating decimal which can be written as an infinite geometric series: $x = 0.037 + 0.000037 + 0.000000037 + .

Vedclass · Accepted Answer

$\frac{37}{999}$

Vedclass · Answer

(D) Let $x = 0.037037037...$ This is a repeating decimal which can be written as an infinite geometric series: $x = 0.037 + 0.000037 + 0.000000037 + ...$ $x = \frac{37}{10^3} + \frac{37}{10^6} + \frac{37}{10^9} + ...$ This is a geometric progression with the first term $a = \frac{37}{1000}$ and common ratio $r = \frac{1}{1000}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$. $x = \frac{37/1000}{1 - 1/1000} = \frac{37/1000}{999/1000} = \frac{37}{999}$. Alternatively,$x = \frac{37}{999} = \frac{1}{27}$. Since $\frac{37}{999}$ is provided as option $D$,the correct answer is $D$.

The value of $\overline{0.037}$,where $\overline{0.037}$ stands for the number $0.037037037...$,is:

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