№1 а6=45 а14=-43 найти s10 № 2 а1=10 s14=1050 найти: d

1) a6=a1+5d=45
a14 = a1+ 13d = -43

a1+13d-a1-5d=-43-45
8d=-88
d=-11
a1=45-5d=45+55=100

S10 = (2a1+9d)/2 * 10 = (2a1+9d)5 = (200-99)*5=101*5=505

2) S14 = (2a1+13d)/2 * 14 = (2a1+13d)*7 = 1050
(20+13d)*7=1050
20+13d = 150
13d=130
d=10