Rozwiązanie:
Zadanie 6.
[tex]a_{3}=16\\a_{5}=12\\a_{5}-a_{3}=12-16=-4=2r\\r=-2\\a_{1}=16-2r=16+4=20\\a_{36}=a_{1}+35r=20-70=-50\\S_{20}=\frac{2*20+19*(-2)}{2}*20=20\\a_{n}=20+(n-1)*(-2)=-2n+22[/tex]
Zadanie 7.
[tex]a_{3}=16\\a_{6}=2\\q^{3}=\frac{a_{6}}{a_{3}}=\frac{2}{16}= \frac{1}{8}\\q=\frac{1}{2}\\a_{1}=\frac{16}{q^{2}}=4*16=64\\a_{10}=a_{1}q^{9}=64*\frac{1}{512} =\frac{1}{8} \\S_{10}=64*\frac{1-\frac{1}{1024} }{1-\frac{1}{2} } =64*\frac{1023}{1024} *2=128*\frac{1023}{1024}=\frac{1023}{8} \\a_{n}=64*(\frac{1}{2} )^{n-1}=(\frac{1}{2} )^{-6}*(\frac{1}{2} )^{n-1}=(\frac{1}{2} )^{n-7}[/tex]
