Which term of the sequence 2, $2\sqrt{2}$, 4, ... is 128?                                                                                                  [Page No. 194]

$$\text{This is a GP with }a=2,r=\frac{2\sqrt{2}}{2}=\sqrt{2}$$

General term: $T_n = ar^{n-1}$

$$T_n = 2(\sqrt{2})^{n-1}$$

Set equal to 128: $2(\sqrt{2})^{n-1} = 128$

$$(\sqrt{2})^{n-1} = 64$$

$$2^{\frac{n-1}{2}} = 2^6$$

$$\frac{n-1}{2} = 6$$

$$n - 1 = 12$$

$$n = 13$$

Therefore, 13th term of sequence is 128.