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I did the following:
$$3.72\overline{4} = 3.724+\left(\left(\frac{4}{10^4}\right)+\left(\frac{4}{10^5}\right)+\left(\frac{4}{10^6}\right)+\cdots\right)$$ where $a=3.724$ and $r=\dfrac{1}{10}$

Using $S=\dfrac{a}{1-r}$ I get $\dfrac{37.24}{9}\neq3.72\overline{4}$

The answer given by the text is $\dfrac{838}{225}$

I've been staring at this for an hour and can't figure out where I'm going wrong.

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  • $\begingroup$ Your first term in the series is $3.724$, but your next term is not $3.724\times\frac{1}{10}$. Not all of your fraction will actually be part of the series. $\endgroup$ – Steven Stadnicki Jul 6 '14 at 18:24
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Start your geometric series after $3.72$: $$ 3.72\overline{4} = 3.72+\left(\left(\frac{4}{10^3}\right)+\left(\frac{4}{10^4}\right)+\left(\frac{4}{10^5}\right)+\cdots\right) $$

Then you have $a=\dfrac{4}{10^3}$ and $r=\dfrac 1{10}$.

The ultimate sum is then

$$ 3.72 + \frac{a}{1-r}. $$

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First of all, your $a$ is incorrect. I think reviewing the formula for a geometric series would be helpful. The way to do these kinds of problems is to split it up into two parts: one involving the part without the repeated pattern (in this case $3.72=\frac{372}{100}$) and the other part involving the repeated part (in this case $.00444\ldots$) which can be rewritten as a geometric series.

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Hint: You consider only the geometric progression, it's $((\frac{4}{10^4})+(\frac{4}{10^5})+(\frac{4}{10^6})+\cdots)$. For this sequence $q=\frac{1}{10}$ and $a=\frac{4}{10^4}$, not $a=3.724$.

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Only part of this is geometric: $$ 3.72\bar{4}= 3.72 +0.00\bar 4=\frac{372}{100}+\frac{4}{1000}\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\cdots\right) $$

$$ =\frac{372}{100}+\frac{4}{1000}\cdot\frac{1}{1-\frac{1}{10}} = \frac{372}{100}+\frac{4}{1000}\cdot\frac{10}{9} =\frac{372}{100}+\frac{4}{900} $$

$$ =\frac{4\cdot93}{4\cdot25}+\frac{4\cdot1}{4\cdot225}=\frac{9\cdot93}{9\cdot25} +\frac{1}{225}=\frac{837}{225}+\frac{1}{225}=\boxed{\dfrac{838}{225}} $$

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