Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
{u_1} + {u_5} = 204\\
{u_2} + {u_6} = 102
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} + {u_1}.{q^4} = 204\\
{u_1}.q + {u_1}{q^5} = 102
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + {q^4}} \right) = 204\\
{u_1}q.\left( {1 + {q^4}} \right) = 102
\end{array} \right.\\
\Rightarrow \frac{{{u_1}\left( {1 + {q^4}} \right)}}{{{u_1}.q.\left( {1 + {q^4}} \right)}} = 2 \Leftrightarrow q = \frac{1}{2} \Rightarrow {u_1} = 192\\
{S_n} = 181\\
\Leftrightarrow {u_1} + {u_2} + {u_3} + .... + {u_n} = 181\\
\Leftrightarrow {u_1} + {u_1}.q + {u_1}.{q^2} + .... + {u_1}.{q^{n - 1}} = 181\\
\Leftrightarrow {u_1}.\left( {1 + q + {q^2} + .... + {q^{n - 1}}} \right) = 181\\
\Leftrightarrow {u_1}.\frac{{{q^n} - 1}}{{q - 1}} = 181\\
\Leftrightarrow 192.\frac{{{{\left( {\frac{1}{2}} \right)}^n} - 1}}{{\frac{1}{2} - 1}} = 181\\
\Rightarrow n = .....\\
b,\\
{u_n} = \frac{3}{{256}} \Leftrightarrow {u_1}.{q^{n - 1}} = \frac{3}{{256}} \Rightarrow n = 15\\
c,\\
{S_n} = {u_1} + {u_2} + {u_3} + .... + {u_n}\\
= {u_1}.\frac{{{q^{n + 1}} - 1}}{{q - 1}} = 192.\frac{{{{\left( {\frac{1}{2}} \right)}^{n + 1}} - 1}}{{\frac{1}{2} - 1}} = 384
\end{array}\)
