anonymous 3 years ago By looking at 0.36 (repeating decimal) like an infinite geometric sequence, convert it into a fraction.

1. anonymous

you can do this a couple ways

2. anonymous

is it $$\overline{.36}$$ ?

3. anonymous

Yes.

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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 $S=\frac{a_1}{1-r}$

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

Thanks.

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

Ah. Well, thanks!

17. anonymous

@erin512 repost i cannot find it