Odpowiedź :
[tex]Zad. 1\\a_1=3\\a_2=k+1\\a_3=3k-6\\\\a_2=\frac{a_1+a_3}2\\k+1=\frac{3+3k-6}2 /*2\\2k+2=3k-3\\2+3=3k-2k\\5=k\\\\\text{Ciag jest arytmetyczny dla k=5}[/tex]
[tex]Zad. 2\\\left \{ {{a_3=-9} \atop {a_{15}=-13}} \right.[/tex]
[tex]\left \{ {{a_1+2r=-9 /*(-1)} \atop {a_1+14r=-13}} \right. \\+\left \{ {{-a_1-2r=9} \atop {a_1+14r=-13}} \right. \\\\12r=-13+9\\12r=-4 /:12\\r=-\frac4{12}\\r=-\frac13\\\\a_1+2*(-\frac13)=-9\\a_1-\frac23=-9 /+\frac23\\a_1=-8\frac33+\frac23\\a_1=-8\frac13\\\\a_n=-8\frac13+(n-1)*(-\frac13)\\a_n=-8\frac13-\frac13n+\frac13\\\underline{a_n=-8-\frac13n}[/tex]
[tex]Zad. 3\\a_n=13-\frac1{10}n\\\\a_{n+1}=13-\frac1{10}(n+1)\\a_{n+1}=13-\frac1{10}n-\frac1{10}\\a_{n+1}=12\frac9{10}-\frac1{10}n\\\\a_{n+2}=13-\frac1{10}(n+2)\\a_{n+2}=13-\frac1{10}n-\frac2{10}\\a_{n+2}=12\frac8{10}-\frac1{10}n\\a_{n+2}=12\frac45-\frac1{10}n[/tex]
[tex]a_{n+1}=\frac{a_n+a_{n+2}}2 /*2\\2a_{n+1}=a_n+a_{n+2}[/tex]
[tex]a_{n+1}=\frac{a_n+a_{n+2}}2 /*2\\2a_{n+1}=a_n+a_{n+2}\\2(\frac{129}{10}-\frac1{10}n)=13-\frac1{10}n+12\frac45-\frac1{10}n\\\frac{129}5-\frac2{10}n=25\frac45-\frac2{10}n\\25.8-\frac15n=25.8-\frac15n\\\\L=P[/tex]
[tex]r=a_{n+1}-a_n\\r=12\frac9{10}-\frac1{10}n-(13-\frac1{10}n)\\r=12\frac9{10}-\frac1{10}n-13+\frac1{10}n\\r=-\frac1{10} \\\\r=a_{n+2}-a_{n+1}\\r=12\frac45-\frac1{10}n-(12\frac9{10}-\frac1{10}n)\\r=12\frac8{10}-\frac1{10}n-12\frac9{10}+\frac1{10}n\\r=-\frac1{10} - \text{Roznica stala}\\\\\\\text{Jest to ciag arytmetyczny}[/tex]
[tex]Zad. 4\\a_1=-5\\a_2=-7\\a_3=-9\\a_n=-31\\\\r=a_2-a_1\\r=-7+5=-2\\\\-31=-5+(n-1)*(-2)\\-31=-5-2n+2\\-31+5-2=-2n\\-28=-2n /:(-2)\\14=n\\\\S_{14}=\frac{-5-31}2*14\\\underline{S_{14}=-36*7=-252}[/tex]