Odpowiedź:
[tex]a)\\\\a_{5}=19\ \ i\ \ a_{9}=35\\\\\\r=\dfrac{a_{9}-a_{5}}{9-5}=\dfrac{35-19}{4}=\dfrac{16}{4}=4\\\\\\a_{n}=a_{1}(n-1)\cdot r\\\\a_{5}=a_{1}+(5-1)\cdot4\\\\19=a_{1}+4\cdot4\\\\19=a_{1}+16\\\\19-16=a_{1}\\\\3=a_{1}\\\\a_{1}=3[/tex]
[tex]b)\\\\a_{3}=-1\ \ i\ \ a_{9}=5\\\\\\r=\dfrac{a_{9}-a_{3}}{9-3}=\dfrac{5-(-1)}{6}=\dfrac{5+1}{6}=\dfrac{6}{6}=1\\\\\\a_{n}=a_{1}+(n-1)\cdot r\\\\a_{3}=a_{1}+(3-1)\cdot1\\\\-1=a_{1}+2\\\\-1-2=a_{1}\\\\-3=a_{1}\\\\a_{1}=-3[/tex]
[tex]c)\\\\a_{3}=-22\ \ i\ \ a_{12}=5\\\\\\r=\dfrac{a_{12}-a_{3}}{12-3}=\dfrac{5-(-22)}{9}=\dfrac{5+22}{9}=\dfrac{27}{9}=3\\\\\\a_{n}=a_{1}+(n-1)\cdot r\\\\a_{3}=a_{1}+(3-1)\cdot3\\\\-22=a_{1}+2\cdot3\\\\-22=a_{1}+6\\\\-22-6=a_{1}\\\\-28=a_{1}\\\\a_{1}=-28[/tex]
