Đáp án:
$\begin{array}{l}
a)S = \frac{{{u_1}}}{{1 - q}}\\
\Rightarrow 4\pi = \frac{{{u_1}}}{{1 - \left( { - \frac{1}{\pi }} \right)}}\\
\Rightarrow {u_1} = 4\pi .\left( {1 + \frac{1}{\pi }} \right) = 4\pi .\frac{{\pi + 1}}{\pi } = 4\pi + 4\\
b)\left\{ \begin{array}{l}
{u_1} + {u_1}.q + {u_1}.{q^2} = \frac{{26}}{{45}}\\
S = \frac{{{u_1}}}{{1 - q}} = \frac{3}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + q + {q^2}} \right) = \frac{{26}}{{45}}\\
{u_1} = \frac{3}{5}.\left( {1 - q} \right)
\end{array} \right.\\
\Rightarrow \frac{3}{5}.\left( {1 - q} \right).\left( {1 + q + {q^2}} \right) = \frac{{26}}{{45}}\\
\Rightarrow 1 - {q^3} = \frac{{26}}{{27}}\\
\Rightarrow {q^3} = \frac{1}{{27}}\\
\Rightarrow q = \frac{1}{3}\\
\Rightarrow {u_1} = \frac{3}{5}.\left( {1 - q} \right) = \frac{2}{5}\\
c)\left\{ \begin{array}{l}
{u_1}.{u_1}.q.{u_1}.{q^2} = - \frac{1}{{27}}\\
\frac{{{u_1}}}{{1 - q}} = \frac{3}{4}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.q = - \frac{1}{3}\\
4{u_1} = 3 - 3q
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
q = - \frac{1}{3}\left( {do:q < 1} \right)\\
{u_1} = 1
\end{array} \right.
\end{array}$
