Odpowiedź:
[tex]a)\\a_3=9\quad a_4=6\\\displaystyle q=\frac{a_4}{a_3} =\frac{6}{9} =\frac{2}{3} \\a_n=a_1q^{n-1}\\a_3=a_1q^2\\9=a_1\left(\frac{2}{3} \right)^2\\\frac{4}{9} a_1=9/\cdot\frac{9}{4} \\a_1=\frac{81}{4} \\\underline{a_n=\frac{81}{4}\cdot \left(\frac{2}{3} \right)^{n-1}}\\b)\\a_3=24\quad a_5=48\\\\\frac{a_5}{a_3} =\frac{a_1q^4}{a_1q^2} =q^{2} =2\quad \Rightarrow \quad q_1=\sqrt{2}\quad \lor \quad q_2=-\sqrt{2} \\a_3=a_1q^2\\24=a_1\cdot2/:2\\a_1=12\\a_n=a_1q^{n-1}\\[/tex]
[tex]a_n=12\cdot(\sqrt{2} )^{n-1}\qquad \lor\quad a_n=12\cdot(-\sqrt{2} )^{n-1}[/tex]
7)
[tex]\displaystyle S_n=a_1\frac{1-q^n}{1-q} \\a_1=4\\q=2\\S_n=4\cdot\frac{1-2^n}{1-2} =4(2^n-1)=4\cdot2^n-4=2^{n+2}-4\\2^{n+2}-4=1020\\2^{n+2}=1024\\2^{n+2}=2^{10}\\n+2=10\\\underline {n=8}[/tex]
Należy zsumować pierwsze 8 wyrazów
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Odpowiedź:
[tex]a)\\a_3=9\quad a_4=6\\\displaystyle q=\frac{a_4}{a_3} =\frac{6}{9} =\frac{2}{3} \\a_n=a_1q^{n-1}\\a_3=a_1q^2\\9=a_1\left(\frac{2}{3} \right)^2\\\frac{4}{9} a_1=9/\cdot\frac{9}{4} \\a_1=\frac{81}{4} \\\underline{a_n=\frac{81}{4}\cdot \left(\frac{2}{3} \right)^{n-1}}\\b)\\a_3=24\quad a_5=48\\\\\frac{a_5}{a_3} =\frac{a_1q^4}{a_1q^2} =q^{2} =2\quad \Rightarrow \quad q_1=\sqrt{2}\quad \lor \quad q_2=-\sqrt{2} \\a_3=a_1q^2\\24=a_1\cdot2/:2\\a_1=12\\a_n=a_1q^{n-1}\\[/tex]
[tex]a_n=12\cdot(\sqrt{2} )^{n-1}\qquad \lor\quad a_n=12\cdot(-\sqrt{2} )^{n-1}[/tex]
7)
[tex]\displaystyle S_n=a_1\frac{1-q^n}{1-q} \\a_1=4\\q=2\\S_n=4\cdot\frac{1-2^n}{1-2} =4(2^n-1)=4\cdot2^n-4=2^{n+2}-4\\2^{n+2}-4=1020\\2^{n+2}=1024\\2^{n+2}=2^{10}\\n+2=10\\\underline {n=8}[/tex]
Należy zsumować pierwsze 8 wyrazów