Mamy

$\{\begin{array}{l}{a}_1+a_3+a_5=21\\ a_3-a_1=3\end{array}$  

Ze wzorów

$\{\begin{array}{l}{a}_1+a_1{q}^2+a_1{q}^4=21\\ a_1{q}^2-a_1=3\end{array}$  

$\{\begin{array}{l}{a}_1\left(1+q^2+q^4\right)=21\\ a_1\left(q^2-1\right)=3\mid :\left(q^2-1\right)\end{array}$  

$\{\begin{array}{l}{a}_1\left(1+q^2+q^4\right)=21\\ a_1=\frac{3}{q^2-1}\end{array}$  

$\{\begin{array}{l}\frac{3}{q^2-1}\cdot \left(1+q^2+q^4\right)=21\mid \cdot \frac{q^2-1}{3}\\ a_1=\frac{3}{q^2-1}\end{array}$  

$\{\begin{array}{l}1+q^2+q^4=7\left(q^2-1\right)\\ a_1=\frac{3}{q^2-1}\end{array}$  

$\{\begin{array}{l}1+q^2+q^4=7q^2-7\mid -7q^2+7\\ a_1=\frac{3}{q^2-1}\end{array}$  

$\{\begin{array}{l}8-6q^2+q^4=0\\ a_1=\frac{3}{q^2-1}\end{array}$  

Do równania

$8-6q^2+q^4=0$  

podstawmy

$t=q^2,t>0$  

Otrzymamy

$t^2-6t+8=0$  

${\Delta }_t={\left(-6\right)}^2-4\cdot 1\cdot 8=36-32=4,\sqrt{{\Delta }_t}=2$  

$t=\frac{6-2}{2}=2\vee t=\frac{6+2}{2}=4$  

Stąd

$q^2=2\vee q^2=4$  

$q=\sqrt{2}\vee q=-\sqrt{2}<0\vee q=2\vee q=-2<0$  

Ciąg jest monotoniczny, więc q > 0. Zatem

$a_1=\frac{3}{{\sqrt{2}}^2-1}=3\vee a_1=\frac{3}{2^2-1}=1$  

Wyraz ogólny

$a_n=3\cdot {\left(\sqrt{2}\right)}^{n-1}$  

lub

$a_n=1\cdot 2^{n-1}$