Bài tập 3.28 trang 131 SBT Toán 11

a) Ta có \(\left\{ \begin{array}{l}
{u_1} + {u_5} = 51\\
{u_2} + {u_6} = 102
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + {u_1}.{q^4} = 51\\
{u_1}.q + {u_1}.{q^5} = 102
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + {u_1}.{q^4} = 51\\
q\left( {{u_1}q + {u_1}{q^4}} \right) = 102
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
q = 2\\
{u_1} = 3
\end{array} \right.\)

b) \({S_n} = \frac{{{u_1}\left( {{q^n} - 1} \right)}}{{q - 1}} = 3069 \Leftrightarrow 3\left( {{2^n} - 1} \right) = 3069 \Leftrightarrow n = 10\)

c) \({u_n} = {u_1}{q^{n - 1}} = 12288 \Leftrightarrow {3.2^{n - 1}} = 12288 \Leftrightarrow n = 12\)

-- Mod Toán 11 HỌC247