Question

数列an是等差数列,bn是等比数列,满足b1=a1^2,b2=a2^2,b3=a3^2,求数列bn公比q

Answer 1

设an公差是d,

因为b1*b3=(b2)^2

所以a1^2*(a1+2d)^2=[(a1+d)^2]^2

(a1^2+2a1d)^2[(a1+d)^2]^2=0

[a1^2+2a1d+(a1+d)^2][a1^2+2a1d-(a1+d)^2]=0

所以a1^2+2a1d+(a1+d)^2=0,

或a1^2+2a1d-(a1+d)^2=0

若a1^2+2a1d-(a1+d)^2=0,则d=0,不合

若a1^2+2a1d+(a1+d)^2=0,

即2a1^2+4a1d+d^2=0

亦即2(a1/d)^2+4a1/d+1=0

解得a1/d=-1±√2/2

若a1/d=-1-√2/2

公比q=b2/b1=(a2/a1)^2=(1+d/a1)^2=(√2-1)^2=3-2√2

若a1/d=-1+√2/2

公比q=b2/b1=(a2/a1)^2=(1+d/a1)^2=（-1-√2）^2=3+2√2