Gọi q là công bội của cấp số. Khi đó ta có
\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{c}}
{{u_1} + {u_2} + {u_3} + {u_4} + {u_5} = 11}\\
{{u_1} + {u_5} = \frac{{82}}{{11}}}
\end{array}} \right.\\
\begin{array}{*{20}{l}}
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{u_2} + {u_3} + {u_4} = \frac{{39}}{{11}}}\\
{{u_1} + {u_5} = \frac{{82}}{{11}}}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{u_1}\left( {q + {q^2} + {q^3}} \right) = \frac{{39}}{{11}}}\\
{{u_1}\left( {1 + {q^4}} \right) = \frac{{82}}{{11}}}
\end{array}} \right.}
\end{array}
\end{array}\)
Suy ra:
\(\begin{array}{l}
\begin{array}{*{20}{l}}
{\frac{{{q^4} + 1}}{{{q^3} + {q^2} + q}} = \frac{{82}}{{39}}}\\
{ \Leftrightarrow 39{q^4} - 82{q^3} - 82{q^2} - 82q + 39 = 0}
\end{array}\\
\begin{array}{*{20}{l}}
{ \Leftrightarrow \left( {3q - 1} \right)\left( {q - 3} \right)\left( {13{q^2} + 16q + 13} \right) = 0}\\
{ \Leftrightarrow q = \frac{1}{3},q = 3}
\end{array}\\
\begin{array}{*{20}{l}}
{q = \frac{1}{3} \Rightarrow {u_1} = \frac{{81}}{{11}}}\\
\begin{array}{l}
\Rightarrow {u_n} = \frac{{81}}{{11}}.\frac{1}{{{3^{n - 1}}}}\\
\begin{array}{*{20}{l}}
{q = 3 \Rightarrow {u_1} = \frac{1}{{11}}}\\
{ \Rightarrow {u_n} = \frac{{{3^{n - 1}}}}{{11}}}
\end{array}
\end{array}
\end{array}
\end{array}\)
