已知等比数列An的前n项和为Sn,公比q＞0,S2=8,S4=80,等差数列Bn满b1=a1,b2+b4=2a2+4,求数列An及数列Bn的通项公式

Question

已知等比数列An的前n项和为Sn,公比q＞0,S2=8,S4=80,等差数列Bn满b1=a1,b2+b4=2a2+4,
求数列An及数列Bn的通项公式

Answer 1

s2=a1+a2
s4=a1+a2+a3+a4
=a1+a2+q^2(a1+a2)
=(1+q^2)(a1+a2)
s4/s2=(1+q^2)(a1+a2)/(a1+a2)
=1+q^2
10=1+q^2
q^2=9(q＞0)
q=3
s2=a1+a2
8=a1+3a1
4a1=8
a1=2
a2=2*3=6
an=a1q^(n-1)=2*3^(n-1)
b1=2
b2+b4=2a2+4
b1+d+b1+3d=2*6+4
2b1+4d=16
2*2+4d=16
4d=12
d=3
bn=b1+(n-1)d
=2+(n-1)*3
=3n-1