在四个数,前三个数成等比数列,其和为19,后三个数为等差数列,其和为15,求此四个数?

1个回答

  • 设这4个数为,a,aq,aq^2,2aq^2-aq,

    则,

    19=a+aq+aq^2,

    15=aq+aq^2+2aq^2-aq=3aq^2,aq^2 = 5.a = 5/q^2,

    19 = a+aq+aq^2 = a+aq+5 = a(1+q)+5=5(1+q)/q^2 + 5,

    0 = 14q^2 - 5q - 5,

    Delta = 5^2 + 4*5*14=25+280=305,

    q = [5+(305)^(1/2)]/28 或 q = [5-(305)^(1/2)]/28.

    q^2 = [330+10(305)^(1/2)]/(28*28)或q^2 = [330-10(305)^(1/2)]/(28*28)

    a = 5/q^2 = 5*28*28/[330+10(305)^(1/2)] = 14*28/[33+(305)^(1/2)]=[33-(305)^(1/2)]/2

    或 a = 14*28/[33-(305)^(1/2)] = [33+(305)^(1/2)]/2

    这4个数为[33-(305)^(1/2)]/2,[(305)^(1/2)-5]/2,5,10-[(305)^(1/2)-5]/2=[25-(305)^(1/2)]/2,

    或,

    [33+(305)^(1/2)]/2,[-(305)^(1/2)-5]/2,5,10+[(305)^(1/2)+5]/2=[25+(305)^(1/2)]/2