Tìm số hạng đầu và công bội

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
{u_5} - {u_1} = 15\\
{u_4} - {u_2} = 6
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}.{q^4} - {u_1} = 15\\
{u_1}.{q^3} - {u_1}.q = 6
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {{q^4} - 1} \right) = 15\\
{u_1}\left( {{q^3} - q} \right) = 6
\end{array} \right.\\
 \Rightarrow \frac{{{q^4} - 1}}{{{q^3} - q}} = \frac{{15}}{6} \Leftrightarrow \frac{{\left( {{q^2} - 1} \right)\left( {{q^2} + 1} \right)}}{{q\left( {{q^2} - 1} \right)}} = \frac{5}{3}\\
 \Leftrightarrow \frac{{{q^2} + 1}}{q} = \frac{5}{2}\\
 \Leftrightarrow 2{q^2} - 5q + 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
q = 2 \Rightarrow {u_1} = 1\\
q = \frac{1}{2} \Rightarrow {u_1} =  - 16
\end{array} \right.\\
b,\\
\left\{ \begin{array}{l}
{u_2} - {u_4} + {u_5} = 10\\
{u_3} - {u_5} + {u_6} = 20
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}q - {u_1}{q^3} + {u_1}.{q^4} = 10\\
{u_1}{q^2} - {u_1}{q^4} + {u_1}.{q^5} = 20
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}q\left( {1 - {q^2} + {q^3}} \right) = 10\\
{u_1}{q^2}\left( {1 - {q^2} + {q^3}} \right) = 20
\end{array} \right.\\
 \Rightarrow \frac{{{u_1}q\left( {1 - {q^2} + {q^3}} \right)}}{{{u_1}{q^2}\left( {1 - {q^2} + {q^3}} \right)}} = \frac{{10}}{{20}}\\
 \Leftrightarrow \frac{1}{q} = \frac{1}{2} \Rightarrow q = 2 \Rightarrow {u_1} = 1
\end{array}\)