设{an}是等差数列{bn}是各项都是正数的等比数列且a1=b1=2,a3+b5=36,a5+b3=14

a3+b5=36.A
a5+b3=14.B
a1+4d+b1q^2=14
2+4d+2q^2=14
4d+2q^2=12
2d+q^2=6
2d=6-q^2.C
B-A
a5-a3+b3-b5=-22
2d+b1q^2-bq^4=-22
2d+2q^2-2q^4+22=0.D
6-q^2+2q^2-2q^4+22=0
-2q^4+q^2+28=0
2q^4-q^2-28=0
(2q^2+7)(q^2-4)=0
q^2=4 {bn}是各项都是正数的等比数列
q=2
d=1
an=a1+(n-1)d=2+n-1=n+1
bn=b1q^(n-1)=2*2^(n-1)=2^n
san=[a1+an]n/2=(2+n+1)/2=(n+3)/2
sTn=b1(1-q^n)/(1-q)=2(1-2^n)/(-1)=2^(n+1)-2