三数成等比数列,其和为7,平方和为99,求这三个数

2个回答

  • -.-

    a1+a2+a3=7

    a1^2+a2^2+a3^2=99

    所以有 a2(1/q+1+q)=7 ①

    a2^2(1/q^2+1+q^2)=99 ②

    (a1+a2+a3)^2=49=a1^2+a2^2+a3^2+2a1a2+2a1a3+2a2a3

    -50=2(a1a2+a1a3+a2a3)

    a1a2+a1a3+a2a3=-25 → a2^2(1+q+1/q)=-25

    ①^2/② → (1+q^2+1/q^2+2/q+2q+2)/(1/q^2+1+q^2)=49/99

    50(q^2+1/q^2)+198(q+1/q)+248=0

    令q+1/q=t 所以 q^2+1/q^2=t^2-2

    所以有 50(t^2-2)+198t+248=0

    50t^2+198t+148=0 t=-1 或者t=-2.96

    q+1/q=-1 (舍) or -2.96

    1+q+1/q=-1.96 a2^2=625/49 a2=25/7 or -25/7

    得到 q=(37+2√186)/25 or (37-2√186)/25

    (a1,a2,a3)=((37-2√186)/7,25/7,(37+2√186)/7)

    or ((37+2√186)/7,25/7,(37-2√186)/7)

    or (-(37+2√186)/7,-25/7,-(37-2√186)/7)

    or (-(37-2√186)/7,-25/7,-(37+2√186)/7)