Hey,

1. If 0<r<1 then;

r-1 is negative
r^n1 is also negative as for 0<r<1, r^n<r

This means t is positive

2. If r<-1

r-1 will be negative
r^n-1 can be positive or negative depending on n (even or odd)

This means t can either be negative or positive


Thus conclusively nothing can be said about t's sign it can be either negative or positive

Now as r is a rational number it can be written as p/q

Substituting in the given equation

t= (p^n-q^n)/(p-q)*q^n-1

This can be expressed finally as p'/q'

Hence, t is rational

Answer: A -I only

t = {(r^n) - 1}/ (r - 1)

we know, that (a^n-b^n)is always divisible by (a-b) => the result of division will always be rational

however, r being -ve or =ve depends on n being odd or even.

hence (A) I only