The second term of a geometric sequence is 2 and the last term is 8. If the common ratio is 2√, how many geometric means are there between t

The second term of a geometric sequence is 2 and the last term is 8. If the common ratio is 2√, how many geometric means are there between the first term and the last term?

$$\small{\text{$ \begin{array}{rcl} t_2 &=& t_1\cdot r^1 \qquad | \qquad t_2 = 2 \quad \text{ and }\quad r = 2^{\frac 12} \\ 2 &=& t_1 \cdot 2^{\frac 12}\\ t_1 &=& 2\cdot 2^{-\frac 12}\\ t_1 &=& 2^{ 1- \frac 12 }\\ \mathbf{t_1} & \mathbf{=} & \mathbf{2^{ \frac 12 }}\\\\\\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} t_n &=& t_1\cdot r^{n-1} \qquad | \qquad t_n = 2^3 \quad \text{ and }\quad t_1 = 2^{ \frac 12 } \quad \text{ and } \quad r = 2^{\frac 12}\\ 2^3 &=& 2^{ \frac 12 } \cdot 2^{\frac 12 \cdot(n-1)} \qquad | \qquad 1^2\\ 2^6 &=& 2^1 \cdot 2^{n-1}\\ 2^6 &=& 2^{1+n-1}\\ 2^6 &=& 2^n\\ 6 &=& n\\ \mathbf{ n } & \mathbf{=} & \mathbf{6} \\ \end{array} $}}$$

We have 6 terms, so between the first term and the 6th term are 4 terms.

$$\small{\text{$ \begin{array}{rcl} t_1 &=& 2^{\frac 12} = \sqrt{2}\\ t_2 &=& 2^{\frac 22} = 2 \\ t_3 &=& 2^{\frac 32} = 2^{\frac 22}\cdot 2^{\frac 12} = 2\cdot \sqrt{2}\\ t_4 &=& 2^{\frac 42} = 2^2 = 4\\ t_5 &=& 2^{\frac 52} = 2^{\frac 42}\cdot 2^{\frac 12} = 2^2\cdot \sqrt{2}= 4\cdot \sqrt{2}\\ t_6 &=& 2^{\frac 62} = 2^3 = 8\\ \end{array} $}}$$

The second term of a geometric sequence is 2 and the last term is 8. If the common ratio is 2√, how many geometric means are there between the first term and the last term?

This is the sequence,

$$2,\;\;2\sqrt2,\;\;4,\;\;4\sqrt2,\;\;8$$

I am not sure what geometric mean means in relation to this question.  :/