Answer 1

Given sequence: 2², 2³, 2⁴,...................
Now,

`2^3/2^2=2, 2^4/2^3=2`

Since `2^3/2^2=2^4/2^3=........=2,` the given sequence is a G.P. with first term, a = 2² = 4 and common ratio, r = 2.
Now, t_n = ar^n-1

∴ t₆ = 4 x (2)⁵ = 4 x 32 = 128