ciąg geometryczny
a_{n+1}=a_nq
q=\frac{a_{n+1}}{a_n}
a_n=\frac{(-5)n}{8}
a_1=\frac{(-5)*1}{8}=-0,625
a_2=\frac{(-5)*2}{8}=-1,25
a_3=\frac{(-5)*3}{8}=-0,875
q=\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{a_3}{a_2}
\frac{a_2}{a_1}=\frac{-1,25}{-0,625}=2
\frac{a_3}{a_2}=\frac{-0,875}{-1,25}=0,7
podany ciąg nie jest ciągiem geometrycznym tylko czy wzór wygląda tak
a_n=\frac{(-5)n}{8} czy tak a_n=\frac{(-5)^n}{8}???
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a_1=-4
q=\frac{1}{2}
a_n=a_1g^{n-1}
a_1=-4
a_2=a_1q^1=-4*(\frac{1}{2})^1=-2
a_3=a_1q^2=-4*(\frac{1}{2})^2=-4*\frac{1}{4}=-1
a_4=a_1q^3=-4*(\frac{1}{2})^3=-4*\frac{1}{8}=-\frac{1}{2}
a_5=a_1q^3=-4*(\frac{1}{2})^4=-4*\frac{1}{16}=-\frac{1}{4}
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a_3=9
a_5=18
a_n=a_1q^{n-1}
a_3=a_1q^2
a_5=a_1q^4
\left \{ {{a_1q^2=9} \atop {a_1q^4=18}} \right.
\left \{ {{a_1=\frac{9}{q^2}} \atop {\frac{9}{q^2}q^4=18}} \right.
\left \{ {{a_1=\frac{9}{q^2}} \atop {9q^2=18}} \right.
\left \{ {{a_1=\frac{9}{q^2}} \atop {q^2=9}} \right.
\left \{ {{a_1=\frac{9}{q^2}} \atop {q=3}} \right.
\left \{ {{a_1=\frac{9}{3^2}} \atop {q=3}} \right.
\left \{ {{a_1=\frac{9}{9}} \atop {q=3}} \right.
\left \{ {{a_1=1} \atop {q=3}} \right.
…