Question

Найдите а1 и q геометрической прогрессии ( а n) . если а1+а4 =30. а 2+а3=10

Answer 1

a+a*q^3=30   a(1+q^3)=30 a(1+q)(1-q+q^2)=30  10(1-q+q^2)/q=30

aq+aq^2=10  aq(1+q)=10

 

1-q+q^2=3q

q^2-4q+1=0

q=2+sqr(3)

q=2-sqrt(3)

a=10/(2+sqrt(3))(3+sqrt(3))=10/(9+5sqrt(3))

a=10/(2-sqrt(3))(3-sqrt(3))=10/(3-sqrt(3))=10*(3+sqrt(3))/6=5*(3+sqrt(3))/3