Odpowiedź:
Należy dodać 10 początkowych wyrazów.
Szczegółowe wyjaśnienie:
[tex]a_1 = -4\\a_2 = -1\\a_3 = 1\\a_4 = 5\\S_{n} = 95\\n = ?\\\\r = a_2 - a_1 = -1-(-4) = -1+4 = 3\\\\\\S_{n} =\frac{2a_1+(n-1)r}{2}\cdot n\\\\95 = \frac{2\cdot(-4) + (n-1)\cdot3}{2}\cdot n\\\\95 = \frac{-8+3n-3}{2}\cdot n\\\\95=\frac{3n-11}{2}\cdot n \ \ /\cdot2\\\\(3n-11)n=95\cdot2\\\\3n^{2}-11n = 190\\\\3n^{2}-11n - 190 = 0\\\\\Delta = (-11)^{2}-4\cdot3\cdot(-190) =121+2280 =2401\\\\\sqrt{\Delta} = \sqrt{2401} = 49\\\\n \in N+\\\\n_1 = \frac{11-49}{6} = -\frac{38}{6} \ \notin D[/tex]
[tex]n_2 = \frac{11+49}{6} = \frac{60}{6} = 10\\\\n = 10\\----[/tex]
