Question

已知数列{an}为等差数列,前n项和为Sn,S2=8,S4=32数列{bn}为等比数列,且b1=a1,b2(a2-a1)

Answer 1

S4-S2=a3+a4=32-8=24

a1+a2=S2=8

4d=(a3+a4)-(a1+a2)=16

d=4

a1+(a1+4)=8

a1=2

故an的通项为an=2+4(n-1)=-2+4n

b1=2

b2*4=b1

q=1/4

故bn的通项为2*(1/4)^(n-1)=8*(1/4)^n