Giải thích các bước giải:
Bài 1:
\({u_n} = {u_1}.{q^{n - 1}}\) là CSN có số hạng đầu \({u_1}\) và công sai bằng \(q\)
a,
\({u_n} = {4.3^n} = {12.3^{n - 1}} \Rightarrow \,\,\,\,\,\,\left( {{u_n}} \right)\) là CSN có \({u_1} = 12;\,\,\,q = 3\)
b,
\(\left( {{u_n}} \right)\) không là CSN.
c,
\({u_n} = - \frac{{{2^n}}}{5} = \left( { - \frac{2}{5}} \right){.2^{n - 1}} \Rightarrow \,\,\,\,\left( {{u_n}} \right)\) là CSN có \({u_1} = - \frac{2}{5};\,\,\,\,q = 2\)
Bài 2:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
{u_4} - {u_2} = 25\\
{u_3} - {u_1} = 50
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.{q^3} - {u_1}.q = 25\\
{u_1}.{q^2} - {u_1} = 50
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}q\left( {{q^2} - 1} \right) = 25\\
{u_1}\left( {{q^2} - 1} \right) = 50
\end{array} \right. \Rightarrow \frac{{{u_1}.q.\left( {{q^2} - 1} \right)}}{{{u_1}.\left( {{q^2} - 1} \right)}} = \frac{{25}}{{50}}\\
\Rightarrow q = \frac{1}{2} \Rightarrow {u_1} = - \frac{{200}}{3}\\
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\\
c,\\
\left\{ \begin{array}{l}
{u_5} - {u_1} = 30\\
{u_4} - {u_2} = 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.{q^4} - {u_1} = 30\\
{u_1}{q^3} - {u_1}q = 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {{q^4} - 1} \right) = 30\\
{u_1}q\left( {{q^2} - 1} \right) = 12
\end{array} \right.\\
\Leftrightarrow \frac{{{u_1}\left( {{q^4} - 1} \right)}}{{{u_1}q\left( {{q^2} - 1} \right)}} = \frac{{30}}{{12}}\\
\Leftrightarrow \frac{{{q^4} - 1}}{{q\left( {{q^2} - 1} \right)}} = \frac{5}{2}\\
\Leftrightarrow \frac{{{q^2} + 1}}{q} = \frac{5}{2}\\
\Leftrightarrow 2{q^2} - 5q + 2 = 0\\
\Leftrightarrow \left( {q - 2} \right)\left( {2q - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
q = 2 \Rightarrow {u_1} = 2\\
q = \frac{1}{2} \Rightarrow {u_1} = - 32
\end{array} \right.
\end{array}\)