(aₙ) - ciąg geometryczny
a₁ = 3, a₄ = 6√2
n-ty wyraz ciągu geometrycznego: [tex]a_n = a_1 \cdot q^{n-1}[/tex], gdzie q to iloraz ciągu
[tex]a_4 = 6\sqrt{2} \ i \ a_4 = a_1 \cdot q^{4-1}= 3 \cdot q^3 \\ Zatem: \\ 3 \cdot q^3 = 6\sqrt{2} \ \ \ |:3 \\ q^3 =2\sqrt{2} \\ q^3 =2^1 \cdot 2^{\frac{1}{2}} \\ q^3 = 2^{1+\frac{1}{2}} \\ q^3 =2^{1\frac{1}{2}} \\ q^3 =2^{\frac{3}{2}} \\ q^3 =(2^{\frac{1}{2}})^3 \\ q = 2^{\frac{1}{2}} \\ q = \sqrt{2}[/tex]
[tex]a_3 = a_1 \cdot q^{3 - 1} = 3 \cdot q^2 = 3 \cdot (\sqrt{2})^2 = 3 \cdot 2 = 6 \\ \underline{a_3 = 6}[/tex]
Odp. a₃ = 6.
