Đáp án:$CSC:{u_1} = 1;{u_{n + 1}} = {u_n} + 1$
Giải thích các bước giải:
Do là cấp số cộng nên: ${u_1} + {u_3} = 2{u_2}$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{u_1}.{u_2}.{u_3} = 6\\
u_1^2 + u_2^2 + u_3^2 = 14
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}}\\
{\left( {{u_1} + {u_3}} \right)^2} - 2{u_1}.{u_3} + u_2^2 = 14
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}}\\
{\left( {2{u_2}} \right)^2} - 2.\frac{6}{{{u_2}}} + u_2^2 = 14
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}}\\
5u_2^2 - \frac{{12}}{{{u_2}}} - 14 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}}\\
5u_2^3 - 14{u_2} - 12 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}}\\
{u_2} = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}.{u_3} = \frac{6}{{{u_2}}} = 3\\
{u_1} + {u_3} = 2{u_2} = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1} = 1\\
{u_3} = 3
\end{array} \right.\\
\Rightarrow d = {u_2} - {u_1} = 1\\
\Rightarrow CSC:{u_1} = 1;{u_{n + 1}} = {u_n} + 1
\end{array}$
