Solving a Repeating Decimal: 0.45454545...

[a_ng116]

If you have a repeating decimal such as 0.45454545... and the question was asking for it to be expressed as an infinite series and find the sum of the series, would it be correct to approach it like this:

$$0.45454545... = \frac {45}{100}+ \frac{45}{10000} + \frac{45}{1000000}...$$

so $$a=\frac{45}{100}$$ and $$r=\frac {1}{100} or 0.01$$

then using $$S_\infty= \frac {a}{1-r}$$

i get: $$= \frac {45/100} {1-0.01} = \frac {5}{11}$$

I think this is right but I'm not sure...any thoughts? Please and thank you.

 

[NateTG]

Works for me.

You can check your answer by, for example, doing the division to see what comes up.

 

[Diane_]

Your answer is certainly right. Remember the Algebra I approach to this problem:

x = .45...

100x = 45.45...

100x - x = 45.45... - .45...

99x = 45

x = 45/99 = 5/11

 

[a_ng116]

Thanks for the help guys, I really appreciate it.