1.
[tex]a_{n} = n^{2}-9n+14\\\\n^{2}-9n+14 < 0\\\\\Delta = b^{2}-4ac = (-9)^{2}-4\cdot1\cdot14 = 81-56 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\n \in N+\\\\n_1 = \frac{-(-9)-5}{2} = \frac{4}{2} = 2\\\\n_2 = \frac{-(-9)+5}{2} = \frac{14}{2} = 7\\\\n \in \ (2;7)\\\\n \in \{3,4,5,6\}[/tex]
Odp. Ten ciąg ma 4 ujemne wyrazy.
2.
[tex]a_{n} = 5n-3\\a_1 = 5\cdot1 - 3 = 2\\a_2 = 5\cdot2 - 3 = 7\\r = a_2 - a_1 = 7-2 = 5\\a_4 = 5\cdot4 - 3 = 17\\a_{20} = 5\cdot20-3 = 97\\\\a_5+a_6 + ... + a_{20} = S_{20}-S_4\\\\S_{20} = \frac{a_1 + a_{20}}{2}\cdot20 = \frac{2+97}{2}\cdot20 = 99\cdot10 = 990\\\\S_4 = \frac{a_1+a_4}{2}\cdot4 = \frac{2+17}{2}\cdot4 = 19\cdot2 = 38\\\\a_5+a_6+...+a_{20} = S_{20}-S_4 = 990-38=952[/tex]
3.
[tex]a)\\q = \frac{10-7\sqrt{2}}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{(10-7\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})} = \frac{30+20\sqrt{2}-21\sqrt{2}-28}{9-8} = 2-\sqrt{2}\\\\b)\\a_1 = \frac{a_3}{q^{2}} = \frac{3-2\sqrt{2}}{(2-\sqrt{2})^{2}} = \frac{3-2\sqrt{2}}{4-4\sqrt{2}+2} = \frac{3-2\sqrt{2}}{6-4\sqrt{2}} = \frac{3-2\sqrt{2}}{2(3-2\sqrt{2})} = \frac{1}{2}\\\\c)\\a_5 = 34-24\sqrt{2}[/tex]
[tex]a_7 = a_5\cdotq^{2} = (34-24\sqrt{2})\cdot(2-\sqrt{2})^{2} =\\= (34-24\sqrt{2})(6-4\sqrt{2})=204-68\sqrt{2}-144\sqrt{2}+192=396-280\sqrt{2}\\\\a_7-a_5 = (396-280\sqrt{2})-(34-24\sqrt{2}) = 362-256\sqrt{2}[/tex]
