Question

By looking at 0.36 (repeating decimal) like an infinite geometric sequence, convert it into a fraction.

Answer 1

you can do this a couple ways

Answer 2

is it \(\overline{.36}\) ?

Answer 3

Yes.

Answer 4

I have the formula \[S _{n}=\frac{ a _{1}-a_{1}r ^{n} }{ 1-r }\]
If that helps.

Answer 5

ok one simple method is to call \(x=\overline{.36}\) so \(100x=36\overline{.36}\)
subtract and get \(100x-x=36\) so
\[99x=36\] and therefore
\[x=\frac{36}{99}=\frac{4}{11}\] but that might not be what you want

Answer 6

@satellite73 hey, when you get a chance can you finish helping me with that problem we started? thanks :)

Answer 7

if you want to use the formula you wrote above, then since you are summing an infinite geometric series omit the \(-a_1r^n\) part and go with
\[S=\frac{a_1}{1-r}\]

Answer 8

Oh. I think I copied the formula down incorrectly...

Answer 9

Cuz I was looking through my notes and wondering how I got the answers with the -a1r^n part

Answer 10

yeah since \(r<1\) you have \(\lim_{n\to \infty}r^n=0\)

Answer 11

Thanks.

Answer 12

in this case you can use
\[a=\frac{36}{100}\] and \[r=\frac{1}{100}\]

Answer 13

For reference, though, does that formula equal\[S_{n}= \frac{ a_{1}(1-r ^{n)} }{ 1-r }\]

Answer 14

you get
\[S=\overline{.36}=\frac{\frac{36}{100}}{1-\frac{1}{100}}\]
\[=\frac{\frac{36}{100}}{\frac{99}{100}}\]
\[=\frac{36}{99}\] etc

Answer 15

no that is a formula for a finite series
infinite series is the one i wrote above

Answer 16

Ah. Well, thanks!

Answer 17

@erin512 repost i cannot find it