Question

How do you find the Nth term in the geometric sequence where the first term is 3 and the fourth term is `6sqrt2`?

Answer 1

`a_N = 3*2^((N-1)/2)`

Explanation:

The general term of a geometric sequence can be written:

`a^n = a r^(n-1)`

where `a` is the initial term and `r` the common ratio.

In our case we are given `a_1 = 3` and `a_4 = 6sqrt(2)`

So we find:

`r^3 = (ar^3)/(ar^0) = a_4/a_1 = (6sqrt(2))/3 = 2sqrt(2) = (sqrt(2))^3`

So (assuming the geometric sequence is Real):

`r = root(3)(sqrt(2)^3) = sqrt(2)`

and

`a_n = 3 * (sqrt(2))^(n-1) = 3 * 2^((n-1)/2)`