यदि `A=1+r^(a)+r^(2a)+r^(2a)+....oo` तक तथा `B=1+r^(b)+r^(2b)+...oo` तक,तो सिध्द कीजिए कि `r=((A-1)/(A))^(1//a)=((B-1)/(B))^(1//b)`

Explanation

`A=1+r^(a)+r^(2a)+…+oo,B=1+r^(b)+r^(2b)+….oo`
`A=(1)/(1-r^(a))व B=(1)/(1-r^(b))अतः1-r^(a)=(1)/(A),1-r^(b)=(1)/(B)`
`rArr " "r^(a)=1-(1)/(A),r^(b)=1-(1)/(B)`
`rArr " "r=((A-1)/(A))^(1//a)वr=((B-1)/(B))^(1//b)`
अतः`" "((A-1)/(A))^(1//a)=((B-1)/(B))^(1//b)`