Question

If `A=1+r^a+r^(2a)+` to `ooa n dB=1+r^b+r^(2b)+oo` , prove that `r=((A-1)/A)^(1//a)=((B-1)/B)^(1//b)`

Answer 1

`A = 1+r^a+r^(2a)+...oo`
It is a G.P. with first term `1` and common ratio `r^a`.
`:. A= 1/(1-r^a)`
`=>1-r^a = 1/A`
`=>r^a = 1-1/A = (A-1)/(A)`
`=> r = ((A-1)/(A))^(1/a)`
Now, `B = 1+r^b+r^(2b)+...oo`
It is a G.P. with first term `1` and common ratio `r^b`.
`:. B= 1/(1-r^b)`
`=>1-r^b = 1/B`
`=>r^b = 1-1/B = (B-1)/(B)`
`=> r = ((B-1)/(B))^(1/b)`
`:. r = ((A-1)/(A))^(1/a) = ((B-1)/(B))^(1/b).`