设等差数列{An}的前n项和为Sn,公比是正数的等比数列{Bn}的前n项和为Tn.已知A1=1,B1=3,A3+B3=17,T3-S3

Question

设等差数列{An}的前n项和为Sn,公比是正数的等比数列{Bn}的前n项和为Tn.已知A1=1,B1=3,A3+B3=17,T3-S3

Answer 1

设{an}公差为d,{bn}公比为q.
a3+b3=a1+2d+b1q^2=1+2d+3q^2=17
T3-S3=b1(q^3-1)/(q-1)-3a1-3d=3(1+q+q^2)-3-3d=12
得关于d,q的二元方程组:
3q^2+2d=16 (1)
3q^2+3q-3d=12 (2)
(1)*3+(2)*2
5q^2+2q-24=0
(q-2)(5q+12)=0
q=2或q=-12/5(公比小于0,不满足题意,舍去)
q=2代入(1)
12+2d=16 d=2
an=a1+(n-1)d=1+(n-1)*2=2n-1
bn=b1q^(n-1)=3*2^(n-1)=(3/2)*2^n
an=2n-1 bn=(3/2)*2^n