Question

Geometric series problem

Answer 1

So the addition of the first three terms is 62,755 and the addition of infinite terms is 440.
I need to find the common ratio

Answer 2

So I wrote two equations:
E1. \[\frac{ a1(r^{3}-1) }{ r-1 }=62755\]
E2. \[S=\frac{ a1 }{ 1-r }=440\]

So from E2 I know that
a1=440-440r (Is it correct until here?

Answer 3

I've seen some people dividing E1 and E2...but why?

Answer 4

because everything cancels if you do that

Answer 5

oh...ok...so I will do that, thanks

Answer 6

So i divided 1 and 2, and got
-(r3-1)=142.625
r=5.2125 ... is that correct?

Answer 7

I'd suggest you solve either one of your two equations for a1 (the first term). This will leave you with an equation in r only; hopefully you could solve that for the common ratio, r. Given that you believe you've found r, try finding the first term (a1). Then check whether your a1 and your r satisfy BOTH of you equations.

Answer 8

Ok, let me check...

Answer 9

Got it!

Answer 10

Cool! Great work! Excellent!

Answer 11

Are you sure the sum of the first 3 terms is 62755 and not 62.755?

For the infinite geometric series to converge |r| must be < 0.

Answer 12

Well...my results are working for 62,755...

What I did

a1(r3-1)(1-r)/(r-1)a1=(62,755)/(440)

-(r3-1)=142.625

r=-141.625^(1/3)=-5.2125...and I think it works...

Can you check it please?

Answer 13

The series should diverge when |r| > 1. You are getting |r| > 1.

If it was 62.755, you will get a nice converging series.