$\text{a)}$  

$\{\begin{array}{l}{a}_1=1\\ a_4=125\end{array}$  

$\{\begin{array}{l}{a}_1=1\\ a_1\cdot q^3=125\end{array}$                       

$\{\begin{array}{l}{a}_1=1\\ q^3=125\end{array}$  

$\{\begin{array}{l}{a}_1=1\\ q=5\end{array}$  

Zatem otrzymujemy:

$a_n=a_1\cdot q^{n-1}$  

$a_n=1\cdot 5^{n-1}$  

$a_n=5^{n-1}$  

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$\text{b)}$  

$\{\begin{array}{l}{a}_2+a_3=3\\ a_1=\frac{1}{4}\end{array}$  

$\{\begin{array}{l}{a}_{1}q+a_1{q}^2=3\\ a_1=\frac{1}{4}\end{array}$  

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

$\{\begin{array}{l}q+q^2=12\\ a_1=\frac{1}{4}\end{array}$  

$\{\begin{array}{l}{q}^2+q-12=0\\ a_1=\frac{1}{4}\end{array}$  

$\Delta =1^2-4\cdot 1\cdot \left(-12\right)=1+48=49$  

$q_1=\frac{-1-7}{2}=\frac{-8}{2}<0$  

$q_2=\frac{-1+7}{2}=\frac{6}{2}=3$  

Wiemy, że jest to ciąg monotoniczny, zatem q>0.

Zatem otrzymujemy:

$a_n=a_1\cdot q^{n-1}$  

$a_n=\frac{1}{4}\cdot 3^{n-1}$  

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$\text{c)}$  

$\{\begin{array}{l}{a}_2\cdot a_4=\frac{1}{9}\\ a_3^2-a_5^2=\frac{8}{81}\end{array}$  

$\{\begin{array}{l}{a}_{1}q\cdot a_1{q}^3=\frac{1}{9}\\ {\left(a_1{q}^2\right)}^2-{\left(a_1{q}^4\right)}^2=\frac{8}{81}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ a_1^2{q}^4-a_1^2{q}^8=\frac{8}{81}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ \frac{1}{9}-a_1^2{q}^4\cdot q^4=\frac{8}{81}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ \frac{1}{9}-\frac{1}{9}{q}^4=\frac{8}{81}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ 1-q^4=\frac{8}{9}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ -q^4=-\frac{1}{9}\end{array}$  

$\{\begin{array}{l}{a}_1^2{q}^4=\frac{1}{9}\\ q^4=\frac{1}{9}\end{array}$  

Ciąg jest monotoniczny, zatem q>0.

$\{\begin{array}{l}{a}_1^2\cdot \frac{1}{9}=\frac{1}{9}\\ q^2=\frac{1}{3}\end{array}$  

$\{\begin{array}{l}{a}_1^2=1\\ q=\frac{1}{\sqrt{3}}\end{array}$  

$\{\begin{array}{l}{a}_1=1\\ q=\frac{\sqrt{3}}{3}\end{array}\vee \{\begin{array}{l}{a}_1=-1\\ q=\frac{\sqrt{3}}{3}\end{array}$  

Wyznaczmy wzór ogólny tego ciągu.

$a_n=a_1\cdot q^{n-1}$  

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