设{an}为等差数列,{bn}为等比数列,a1=b1=1,a2+a4=b3,b2b4=a3,分别求出{an}及{bn}的

a1=b1=1
a2=1+d
a4=1+3d
a3=1+2d
b3=q^2
b2=q
b4=q^3
a2+a4=b3
所以1+d+1+3d=q^2,
2+4d=q^2
b2b4=a3
q^4=1+2d
相除
(2+4d)/(1+2d)=q^2/q^4
q^2=1/2
d=(q^2-2)/4=-3/8
q=±√2/2
S10=(a1+a10)*10/2
=(a1+a1+9d)*10/2
=(2-27/8)*5
=-55/8
T10=b1*(1-q^10)/(1-q)
=1*[1-(1/2)^5]/(1±√2/2)
=(62±31√2)/32