an是 等差数列,sn是前n项和,bn等比数列a1= b1=2,a4+b4=27,s4-b4=10 求2个通项

a(n)=2+(n-1)d.
s(n)=2n+n(n-1)d/2.
b(n)=2q^(n-1).
10=s(4)-b(4)=8+6d-2q^3,
27=a(4)+b(4)=2+3d+2q^3,
37=10+9d,d=3.
a(n)=2+3(n-1)=3n-1.
10=8+6d-2q^3=26-2q^3,
q^3=8,q=2.
b(n)=2*2^(n-1)=2^n
t(n)=a(n)b(1)+a(n-1)b(2)+...+a(1)b(n)=(3n-1)*2 + (3n-4)*2^2 + (3n-7)*2^3 + ...+8*2^(n-2)+5*2^(n-1) + 2*2^n,
2t(n)=(3n-1)*2^2 + (3n-4)*2^3 + (3n-7)*2^4 + ...+ 8*2^(n-1) + 5*2^n + 2*2^(n+1),
t(n)=2t(n)-t(n)=-(3n-1)*2 + 3[2^2 + 2^3 + ...+ 2^n] + 2^(n+2)
=2^(n+2) - 2(3n-1) + 12[1+2+...+2^(n-2)]
=2^(n+2)-2(3n-1)+12[2^(n-1)-1]
=2*2^(n+1)-6n+2 +3*2^(n+1)-12
=5*2^(n+1) - 6n - 10
-2a(n)+10b(n)=-2(3n-1)+10*2^n=5*2^(n+1) - 6n + 2 = t(n) + 12

已知｛an｝是等差数列，其前n项为sn，｛bn}是等比数列，且a1=b1=2，a4+b4=27，s4-b4=10，求数列｛an}｛bn｝ a4+b4=2+3d+3q^3=27

a4+b4=27,s4-b4=10
s(4)+a(4)=37
4(a(1)+a(4))/2+a(4)=37
5a(1)+9d=37
a(1)=2
d=3
a(n)=a(1)+d(n-1)=2+3(n-1)=3n-1
b(4)=27-a(1)-3d=27-2-9=16=b(1)q^(4-1)=2*q^3
q=2
b(n)=b(1)q^(n-1)=2*2^(n-1)=2^n