Answer: (a) {a, ar, ar², ar³, ar⁴, ar⁵...}, (b) arⁿ⁻¹
For part (a), the question gives us the first term a, and then asks us to apply the common ratio r six times.
In order for ar = a, the nth term of r will have to equal 0 (this implies that n is an exponent; thus giving us the first term a, as r = 1).
Since we use this method on the first term, we must use it for the next five, in which r gains an additional exponent for every consecutive value (nth term) thereafter.
Ultimately getting: {a, ar, ar², ar³, ar⁴, ar⁵...}
For part (b), we first have to understand that the sequence does not start at 0, but at 1 for n. In order for ar = a, with n = 1, there needs to be subtraction of -1 within the exponent. So that arⁿ⁻¹
If we check and apply this, we can see that:
{ar¹⁻¹, ar²⁻¹, ar³⁻¹, ar⁴⁻¹, ar⁵⁻¹, ar⁶⁻¹...} = {a, ar, ar², ar³, ar⁴, ar⁵...} = arⁿ⁻¹ = Tn
as
. But this cancellation is only valid when 
.
, you end up with the indeterminate form 
, which is why 
is not a zero.
