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//a)`

Answer 1

By summing the given infinite geometric series, we get
`x=1/((1-r^(a))) and y=1/((1-r^(b)))`
`rArr (1-r^(a))=1/x` and `(1-r^(b))=1/y`
`rArr r^(a)=(1-1/x) and r^(b)=(1-1/y)`
`rArr r=((x-1)/x)^(1/a) and r=((y-1)/y)^(1/b)`.
Hence, `r=((x-1)/x)^(1/a)=((y-1)/y)^(1/b)`.