Ciągi dany jest ciąg arytmetyczny pierwszym wyrazie 7 i różnicy 9
Pomoże ktoś z 1,2,3 i4

Odpowiedź:
zadanie 1.
[tex]\\a_{1} = 7\\ r = 9\\a_{5} = a_1 + 4r =7 + 4 * 9=7+36=43\\a_{6} = a_{5} + r = 43 + 9 = 52\\S = a_5 + a_6 = 43 + 52 =95[/tex]
zadanie 2.
[tex]a_7 = 13\\a_{13} = 7\\ \\a_7 + 6r = a_{13}\\13 + 6r = 7\\6r = 7 -13 \\6r= -6\\r= -1\\\\a_1 = a7 -6r\\a_1=13 -6*(-1)\\a_1 = 13+6=19\\\\a_n = a_1 +(n-1)*r\\a_n = 19+(n-1)*(-1)[/tex]
zadanie 3.
[tex]S = 3+10 + 17 +...+346\\a_1 = 3\\a_2=10\\a_3=17\\r=a_2- a_1\\r=10-3=7\\r=7\\\\a_n = 3+(n-1)*r\\a_n = 3+(n-1)*7\\a_n = 3+7n-7\\\\346 = 3-7+7n\\346 = -4+7n\\346+4=7n\\350 = 7n\\ 7n = 350 \\n= 50\\a_{50}=346\\S_n=\frac{a_1+a_n}{2}*n\\S_{50}= \frac{3+346}{2}*50 = \frac{349}{2}*50 = \frac{349}{1} *25\\S_{50} = 349*25=8725[/tex]
zadanie 4.
[tex]a_{15} = 18\\a_{18} = -5\frac{1}{3}\\a_{18} = a_{15}*q^{3} \\-\frac{16}{3} = 18 * q^{3}\\\\q^{3} =-\frac{16}{3}*\frac{1}{18}\\\\ q^{3} = -\frac{8}{3}* \frac{1}{9}\\\\q^{3} = -\frac{8}{27} \\\\q=-\frac{2}{3}[/tex]