Answer: (a) Ā {a, ar, arĀ², arĀ³, arā“, arāµ...}, (b) Ā arāæā»Ā¹
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. Ā
Ultimately getting: {a, ar, arĀ², arĀ³, arā“, arāµ...}
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 arāæā»Ā¹
If we check and apply this, we can see that:
{arĀ¹ā»Ā¹, arĀ²ā»Ā¹, arĀ³ā»Ā¹, arā“ā»Ā¹, arāµā»Ā¹, arā¶ā»Ā¹...} = {a, ar, arĀ², arĀ³, arā“, arāµ...} = arāæā»Ā¹ = Tn
