Wiemy, że:

$a_{10}=2\cdot a_5$  

$a_{10}=a_{15}-2$  

Ze wzoru ogólnego ciągu:

$a_{10}=a_1+9r$  

$a_5=a_1+4r$  

$a_{15}=a_1+14r$  

Możemy więc zapisać:

$\{\begin{array}{l}{a}_1+9r=2\cdot \left(a_1+4r\right)\\ a_1+9r=a_1+14r-2\end{array}$  

$\{\begin{array}{l}{a}_1+9r=2a_1+8r\\ a_1+9r=a_1+14r-2\end{array}$  

$\{\begin{array}{l}{a}_1+9r=2a_1+8r\\ 2a_1+8r=a_1+14r-2\mid -a_1\end{array}$  

$\{\begin{array}{l}{a}_1+9r=2a_1+8r\\ a_1+8r=14r-2\mid -8r\end{array}$  

$\{\begin{array}{l}{a}_1+9r=2a_1+8r\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}6r-2+9r=2\cdot \left(6r-2\right)+8r\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}15r-2=12r-4+8r\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}15r-2=20r-4\mid -15r\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}-2=5r-4\mid +4\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}2=5r\mid :5\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}\frac{2}{5}=r\\ a_1=6r-2\end{array}$  

$\{\begin{array}{l}r=\frac{2}{5}\\ a_1=6\cdot \frac{2}{5}-2\end{array}$  

$\{\begin{array}{l}r=\frac{2}{5}\\ a_1=\frac{12}{5}-2\end{array}$  

$\{\begin{array}{l}r=\frac{2}{5}\\ a_1=\frac{2}{5}\end{array}$  

$\{\begin{array}{l}{a}_1=\frac{2}{5}\\ r=\frac{2}{5}\end{array}$