Diketahui barisan \( 2\sqrt{2}, 4, 4\sqrt{2}, 8, \cdots \). Suku keberapakah 128?
- 11
- 12
- 13
- 14
- 15
Pembahasan:
Dari barisan \( 2\sqrt{2}, 4, 4\sqrt{2}, 8, \cdots \) dapat kita peroleh:
\begin{aligned} r = \frac{U_n}{U_{n-1}} &= \frac{U_3}{U_2} = \frac{4\sqrt{2}}{4} = \sqrt{2} \\[8pt] U_n = ar^{n-1} &\Leftrightarrow 128 = 2\sqrt{2} \cdot (\sqrt{2})^{n-1} \\[8pt] &\Leftrightarrow 128 = 2\sqrt{2} \cdot \frac{(\sqrt{2})^n}{\sqrt{2}} = 2(\sqrt{2})^n \\[8pt] &\Leftrightarrow 64 = (\sqrt{2})^n \Rightarrow 64 = \left( 2^{\frac{1}{2}} \right)^n \\[8pt] &\Leftrightarrow 2^6 = 2^{\frac{1}{2}n} \Rightarrow 6 = \frac{1}{2}n \\[8pt] &\Leftrightarrow n = 12 \end{aligned}
Jawaban B.