Answer 1

Let a₁, a₂, a₃, ................., a_n, ..........be a G.P. with common ratio r.

⇒ `(a_(n+1))/a_n` = r for all n ∈ N

If each term of a G.P. is raised to the power x, we get the sequence  `"a"_1^x,  "a"_2^x,  "a"_3^x, ............,"a"_n^x,.........`

Now, `(a_(n+1))^x/(a_n)^x=((a_(n+1))/a_n)^x=r^x` for all n ∈ N

Hence, `"a"_1^x,  "a"_2^x,  "a"_3^x, ............,"a"_n^x,.........` is also a G.P.