Question

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

Answer 1

a1=b1=1

a2=1+d

a4=1+3d

a3=1+2d

b3=q^2

b2=q

b4=q^3

所以1+d+1+3d=q^2,2+4d=q^2

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