$\{\begin{array}{l}{a}_1+a_3=52\\ a_2^2=100\end{array}$  

$\{\begin{array}{l}{a}_1+a_1\cdot q^2=52\\ {\left(a_1\cdot q\right)}^2=100\end{array}$  

$\{\begin{array}{l}{a}_1\left(1+q^2\right)=52\mid :\left(1+q^2\right)\\ {\left(a_1\cdot q\right)}^2=100\end{array}$  

$\{\begin{array}{l}{a}_1=\frac{52}{1+q^2}\\ {\left(\frac{52}{1+q^2}\cdot q\right)}^2=100\end{array}$  

Rozwiążmy drugie równanie.

${\left(\frac{52q}{1+q^2}\right)}^2=100$  

$\frac{2704q^2}{{\left(1+q^2\right)}^2}=100$  

$2704q^2=100{\left(1+q^2\right)}^2$  

Wstawmy  $t=q^2$  

$2704t=100{\left(1+t\right)}^2$  

$2704t=100\left(1+2t+t^2\right)$  

$2704t=100+200t+100t^2\mid -2704t$  

$0=100t^2-2504t+100\mid :4$  

$0=25t^2-626t+25$  

$\Delta ={\left(-626\right)}^2-4\cdot 25\cdot 25=391876-2500=389376$  

$\sqrt{\Delta }=624$  

$t_1=\frac{626-624}{2\cdot 25}=\frac{2}{50}=\frac{1}{25}$  

$t_2=\frac{626+624}{2\cdot 25}=\frac{1250}{50}=25$  

$\frac{1}{25}=q^2\text{lub}25=q^2$  

$q_1=\frac{1}{5}{q}_2=-\frac{1}{5}{q}_3=5q_4=-5$  

 

I przypadek.

Gdy  $q=\frac{1}{5}$  

$a_1=\frac{52}{1+q^2}$   

$a_1=\frac{52}{1+\frac{1}{25}}$  

$a_1=\frac{52}{\frac{26}{25}}$  

$a_1=52\cdot \frac{25}{26}$  

$a_1=50$  

$a_n=a_1\cdot q^{n-1}$  

$a_n=50\cdot {\left(\frac{1}{5}\right)}^{n-1}$  

$a_n=50\cdot {\left(\frac{1}{5}\right)}^n\cdot {\left(\frac{1}{5}\right)}^{-1}$  

$a_n=50\cdot {\left(\frac{1}{5}\right)}^n\cdot 5$  

$a_n=250\cdot {\left(\frac{1}{5}\right)}^2$  

 

II przypadek.

Gdy  $q=-\frac{1}{5}$  

$a_1=50$  

$a_n=a_1\cdot q^{n-1}$  

$a_n=50\cdot {\left(-\frac{1}{5}\right)}^{n-1}$  

$a_n=50\cdot {\left(-\frac{1}{5}\right)}^n\cdot {\left(-\frac{1}{5}\right)}^{-1}$  

$a_n=50\cdot {\left(-\frac{1}{5}\right)}^n\cdot \left(-5\right)$  

$a_n=-250\cdot {\left(-\frac{1}{5}\right)}^n$  

 

III przypadek.

Gdy  $q=5$  

$a_1=\frac{52}{1+q^2}$  

$a_1=\frac{52}{1+25}$  

$a_1=\frac{52}{26}$  

$a_1=2$  

$a_n=a_1\cdot q^{n-1}$  

$a_n=2\cdot 5^{n-1}$  

$a_n=2\cdot 5^n\cdot 5^{-1}$  

$a_n=\frac{2}{5}\cdot 5^n$  

 

IV przypadek.

Gdy  $q=-5$  

$a_1=2$  

$a_n=a_1\cdot q^{n-1}$  

$a_n=2\cdot {\left(-5\right)}^{n-1}$  

$a_n=2\cdot {\left(-5\right)}^n\cdot {\left(-5\right)}^{-1}$  

$a_n=-\frac{2}{5}\cdot {\left(-5\right)}^n$