Geometric Sequence

Question

We've been doing geometric sequence in school lately and when I was studying, I came across these two problems:

1. Show that 4^(1/6) = (cube root of) 2 using geometric series.
2. Use the sequence: 1, ___, _____, x^5,____. Find the missing terms if the sequence is geometric.

I tried these multiple times and was unable to get the right answer. Can someone please explain them to me? Thank you. :)

Answer 1

2)

x⁰, __, __, x⁵

x⁵ = x⁰ * r³

x^5/3 = r

x⁰, x^5/3, x^10/3, x^15/3, x^20/3

Answer 2

re write 4 as (2)^2 so you now have (2^2)^1/6 = 2^1/3

By the rules of exponents when you raise a power to a power you multiply so this becomes

2^1/3 = 2^1/3 and there you go

Answer 3

4^1/6 = (2^2)^1/6 = 2^2/6 = 2^1/3 = cube root of 2

BUT you asked for proof by geometric sequence?

let the common ratio, r = 2^1/6, then

the sequence is 1, 2^1/6, 2^2/6= 2^1/3, 2^3/6 = 2^1/2, 2^4/6 = 2^2/3, 2^5/6 2^6/6 = 2 or

1, 2^1/6, 2^1/3, 2^1/2, 2^2/3, 2^5/6, 2 ....

the 3rd term is 2^2/6 = 2^1/3 = cube root of 2, but 2^2/6 also = (2^2)^1/6 = 4^1/6 = 2^1/3

r= x^5/3

AN= r^N-1

AN= (x^5/3)^N-1

AN =x^(5N/3 - 5/3)

A1 = x^0 = 1

A2 = x^5/3

A3 = x^10/3

A4 = x^15/3 = x^5